INTRODUCTION
Redox reactions are the chemical reactions in which reduction and oxidation occurs
simultaneously. We can simply define oxidation as the process in which addition of oxygen (or
any electronegative element) or removal of hydrogen (or any electropositive element) takes place
and reduction is vice – versa.
ELECTRONIC CONCEPT OF OXIDATION AND REDUCTION
Oxidation is a process in which an atom or an ion loses one or more electrons
Na
Na e ; Mg Mg2 2e ; MnO24 MnO4 e
Reduction is a process in which an atom or an ion gains one or more electrons
Fe3+ e
Fe2
S
2e
S2
Fe(CN)6
3
e
Fe CN 6
4
REDOX REACTION
Redox reactions may be regarded as electron transfer reactions in which the electrons are
transferred from one reactant to the other. The substance that loses electron is called a reducing
agent or a reductant while the other which accepts the electron is called on oxidizing agent or
oxidant
oxidation
Zn(s) 2H (aq) Zn2 (aq) H2 (g)
Re ducing agent
Reduction
oxidation
Al Fe 2O3 Al2 O3 2Fe
Re ducing agent oxidi sin gagent
Reduc t ion
REDOX REACTIONS IN AQUEOUS SOLUTION
In aqueous solutions, the spontaneous redox reactions can be carried out directly as well as
indirectly.
Redox Reactions: Redox reactions in which oxidation and reduction takes place in the same
vessel are called redox reactions. In such reactions, the transfer of electrons from reductant to
oxidant occurs over a very short distance (within molecular diameters). For example, if a zinc rod
is placed in a solution of copper sulphate in a beaker, a spontaneous reaction occurs and
following changes will be observed.
Zinc rod
CuSO 4 solution
Zinc rod is eaten away
(forming ZnSO4)
Precipitate of copper
(a)
(b)
P-2123-CBSE-P1-CHEMISTRY-REDOX REACTION & STOICHIOMETRY
2
Observations for redox reaction occurring in a beaker
Zinc rod starts dissolving and loses its mass gradually.
The blue color of the solution starts fading.
Copper metal either starts settling at the bottom of the beaker or depositing on the zinc
rod.
The reaction is exothermic i.e., it takes place with the evolution of heat.
The solution remains electrically neutral throughout.
The reaction will not continue indefinitely but stops after some time.
The overall reaction taking place in the beaker may be represented as:
2
2
Zn(s) Cu(aq)
SO 4(aq)
Zn(aq)
SO24(aq)
Cu(s)
2–
Discarding the common SO4 ions
Zn(s)+ Cu2+ (aq) Zn2+ (aq) + Cu(s) ................
(i)
2+
Zinc loses electrons and changes to Zn ions which go into the solution. As a result, the mass of
2+
zinc rod decreases. The electrons lost by zinc rod are gained by Cu (aq) ions and they change
into Cu(s) atoms which settle down at the bottom of beaker in the form of precipitate. It may be
noted that in the direct redox reaction, the chemical energy appears as heat.
Similarly, when we dip a copper strip in a silver nitrate solution,
copper gets oxidised and go into the solution whereas Ag+
ions accept electrons and get reduced. The reaction may be
written as: Cu(s) + 2Ag+ (aq) 2Ag(s) + Cu2+ (aq) ...(ii)
loss of 2e- : oxidation
Cu(s) 2Ag(aq)
2
Cu(aq)
Copper rod
AgNO 3 solution
2Ag(s)
.... (ii)
Gain of 2e- : Reduction
Thus, copper is oxidised to Cu2+ and Ag+ is reduced to Ag(s). Therefore, copper acts as reducing
agent or reductant while silver acts as oxidising agent or oxidant in equation (ii) and copper acts
as an oxidising agent in equation (i). The reason being that electrondonating ability of zinc is
more than that of Cu while electron donating ability of Cu is more than that of Ag.
Oxidation and reduction half reactions
Every redox reaction can be split up into two half – reactions one representing loss of electron i.e.
oxidation half reaction while the other representing gain of electrons i.e. reduction half reaction
like
Zn Cu2
Zn2 Cu
Zn
Zn2 2e (oxidation half reaction)
Cu+2 2e
2
Cu (Re duction half reaction)
2
Sn4 Hg22 can spilt up into half reactions, as
Sn 2Hg
Sn2
Sn4 2e (oxidation half reaction)
Hg22 (Re duction half reaction)
2Hg2+ 2e
(
)
aq → 2Al
) (
)
)
3+
(
(
(
2+
)
aq
s +3Cu
(
)
(ii) 2Al
)
(
Exercise 1.
Write the half reactions for the following redox reaction
(i) Al +3Ag + → Al 3 + aq + 3Ag s
aq + 3Cu
s
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
3
OXIDATION NUMBER OR OXIDATION STATE
Oxidation no. of an element may be defined as the charge which an atom of the element has in
its ion or appears to have when present in the combined states with other atoms
Rules for assigning oxidation number
The O.N of the element in the free or elementary state is always zero irrespective of its
allotropic form
Oxidation number of He = 0
Oxidation number of Cl2 0
Oxidation number of sulphur in S8 = 0
Oxidation number of phosphorus in P4 0
The O. N. of the element in mono atomic ion is equal to the charge on the ion. For
example, in K Cl , the O.N. of K is +1 while that Cl is –1
O.N. of all alkali metals is +1 while those of alkaline earth metals is +2 in all their
compounds
The O.N. of fluorine is always –1 in all its compounds. (Note that fluorine is most
electronegative element, hence cannot attain positive oxidation state)
Hydrogen is assigned +1 O.N. when it is attached to more electronegative element but it
has –1 O.N. in metal hydrides like NaH,MgH2 ,CaH2 ,LiH etc.
Oxygen is assigned O.N –2 in most of its compounds.
Exceptions: F2O, hence O is in +2 oxidation state.
1
In super oxides like KO2 it has O.N.
2
In peroxides like H2 O2 ,BaO2 ,Na2 O2 etc. O.N is –1.
In ozonide the oxidation number of oxygen is assigned 1/3.
In accordance with principle of conservation of charge, the algebric sum of the oxidation
number of all the atoms in molecule is zero. But in case of polyatomic ion the sum of O.N.
of all its atoms is equal to charge on the ion.
Maximum O.N. of an element can be equal to its group number. (Except O and F)
Minimum O.N. of an element can be equal to (8-n) n is the group no. of the element
(Except metals)
The evaluation of O.N cannot be made directly in some cases e.g. HCN by using the
above rule since we have no rule for oxidation No. of both N and C.
In all such cases evaluation of O.N should be made by taking contribution of covalent
bonds and coordinate bonds
(a) When two atoms are attached with the help of single covalent bond then its
contribution for less electronegative atom is +1 and for more electronegative atom
is –1.
(b) When covalent bond is present between the two same atoms then its contribution will
be zero for both the atoms
(c) In case of Coordinate bond, it gives +2 value of oxidation number to less
electronegative atom and -2 values to more electronegative atom when coordinate
bond is directed from less electronegative atom to more electronegative atom.
(d) If coordinate bond is directed from more electronegative to less electronegative atom
then its contribution will be zero for both the atoms.
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4
Illustration 1. Calculate the O.N. of the star marked elements in the following species/
molecules.
*
*
*
*
Cr2 O72 ,MnO24 ,S 2 O32 ,Cr O5
Solution:
Cr in Cr2O72-
(i)
Let the O.N of Cr be x
O.N. of each O atom = 2
Sum of O.N. of all atoms = 2x +7 (-2) = x – 14
It is the charged species hence the total sum of oxidation numbers = charge
present on the species
2x -14 = -2 or x = +6.
(ii) Mn in MnO24
Let the O.N. of Mn be x
O.N of each O atom = - 2
Sum of O.N. of all atoms will be x + 4(-2) = x -8
Sum must be equal to -2
x – 8 = -2
x = +6
(iii) S in S2O32
Let the O.N. of S be x
O.N. of each O atom = -2
Sum of O.N. of each atom = 2x + 3(-2) = 2x – 6
Sum must be equal to charge i.e. -2
2x – 6 = -2
2 6
x=
2
2
(iv) Cr in CrO5 On calculating the O.N. of Cr in CrO5 by above method is coming
out to be +10 but Cr can show maximum O.N. upto +6 (because group of Cr
is VIB)
We have to see the structure
1
1
O
O
6
Peroxy linkge
Cr
O
O
O
1
2
1
In this structure two peroxy linkages are present due to which four O atoms
shows the O.N.= -1
Let the O.N. of Cr be x
x +(4 -1) + (-2) = 0
x – 4 –2 = 0 x = +6
Exercise 2.
Identify oxidant and reductant in the following redox reactions
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
aq
)
(
(
)
(
)
)
)
(
(
)
(
)
(
)
)
(
(
(
2 Fe
)
+
4
3 CN 6 aq +H 2 O 2 aq + 2H aq 2 Fe CN 6 aq + 2H 2 O
(ii) 2NaClO
2NaIO
3 aq +I 2 aq
3 aq + Cl 2 g
(i)
5
Exercise 3.
Calculate, the O.N. of the star marked elements in the following species/ Molecules
*
*
*
*
Na 2 S 4 O6 ,K M nO 4 ,H 2 S O 5 ,CH 2 Cl 2
BALANCING OF REDOX REACTIONS
Some examples of the redox reactions are
Sn2+ + 2Hg2+ Hg 22 +Sn4+
R
O
MnO 4 +5Fe2+ + 8H+ 5Fe3+ + Mn2+ + 4H2O
O
R
Cr2 O 72 + 6Fe2+ + 14H+ 6Fe3+ + 2Cr3+ + 7H2O
O
R
3Cu + 2NO 3 + 8H+ 2NO + 3Cu2+ + 4H2O
R
O
If one of the half reactions does not take place, other half will also not take place. We can
say oxidation and reduction go side by side.
3Cl2 + 6OH– 5Cl– + ClO 3 + 3H2O
O.N. = 0
-1
+5
R
O
In this we find that Cl2 has been oxidized as well as reduced. Such type of redox reaction
is called Disproportionation reaction. Examples are
+
Disproport ionates
2Cu
Cu + Cu2+
R
O
2HCHO + OH– CH3OH + HCOO –
R
O
4ClO 3 3ClO 4 + Cl–
O
R
3MnO 24
+ 2H2O MnO2 + 2MnO 4 + 4OH–
R
O
We know that during redox reactions there is change in O.N. of the elements due to transfer of
electrons. The number of electrons lost during oxidation is equal to the number of electrons
gained during reduction. It is the basic principle of balancing of redox equations.
The two methods which are frequently employed are O.N. method and ion electron method.
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6
Oxidation number method
Step – 1.
Step – 2.
Step – 3.
Step – 4.
Step – 5.
Step – 6.
Step – 7.
Write the skeletal equation of all the reactants and products of the reaction.
Indicate the oxidation number of each element above its symbol and identify
the elements which undergo a change in the oxidation number (O.N.).
Calculate the increase or decrease in O.N. and identify the oxidizing and
reducing agents. If more than one atom of the same element is involved.
Find out total increase or decrease in O.N. by multiplying this increase or
decrease in O.N per atom by the number of atoms undergoing that change.
Multiplying the formula of the oxidizing and the reducing agents by suitable
integers so as to equalize the total increase or decrease in O.N. as
calculated in step 3.
Balance all atoms other than H and O.
Finally balance H and O atoms by adding H2O molecules using hit and trial
method.
In case of ionic reactions
(a) For acidic medium – First balance O atoms by adding H2O molecules on
O deficient side and then balance H atom by adding H ions to whatever
side deficient in H atoms.
(b) For basic medium –
First balance O atom by adding H2O molecules on O excess side and
then twice OH ions on opposite side.
Illustration 2. Balance the equation K 2Cr2 O7 + HCl
Solution:
KCl + CrCl3 + H2 O + Cl2
Remember that alkali and alkaline earth metals have only one oxidation number
and as long as they remain in the compound they do not undergo oxidation or
reduction.
6
3
1
0
K 2Cr2 O7 + HCl KCl + CrCl3 + H2 O Cl2
Thus here Cr of K 2 Cr2 O7 is reduced to CrCl3 (+6 +3) and Cl of HCl is oxidized
to Cl2 (–1 0). In short.
2Cl-1
Oxidation:
6
2
Reduction:
Cl02 +2e-
-
Cr +6e 2Cr
-1
0
2
+3
... (a)
... (b)
-
Cr2+6 + 6e - 2Cr 3+
Step (iii)
2Cl
Step (iv)
Multiply equation (a) by 6 and (b) by 2
12Cl-1 6Cl02 + 12e -
Cl + 2e ;
2Cr2+6 + 12e- 4Cr +3
2Cr26 12Cl1 4Cr 3 6Cl02
or
Step (v)
Step (vi)
Cr2+6 + 6Cl- 2Cr +3 + 3Cl02
K 2Cr2 O7 + 6HCl
2CrCl3 + 3Cl2
Making provision of KCl and H2O in the product: Since the
reactant has 7 oxygen atoms, in the product 7H2 O must be
present. For accounting 14 hydrogen atoms of water in the
product, the reactants must have 12 HCl molecules (the only H
containing species). For accounting the 2K atoms and
14 – 12 = 2 additional Cl atoms in the reactant, the product must
have 2KCl. Hence the balanced equation is.
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
7
K 2Cr2 O7 + 14HCl
2KCl + 2CrCl3 + 7H2 O + 3Cl2
ION ELECTRON METHOD
This method involves the following steps:
Divide the complete equation into two half reactions, one representing oxidation and the
other reduction.
Balance the atoms in each half reaction separately according to the following steps:
(a) First of all balance the atoms other than H and O.
(b) In a reaction taking place in acidic or neutral medium, oxygen atoms are balanced by
adding molecules of water to the side deficient in oxygen atoms while hydrogen
+
atoms are balanced by adding H ions to the other side deficient in hydrogen atoms.
On the other hand, in alkaline medium (OH-), for every excess of oxygen atom on
one side is balanced by adding one H2O to the same side and 2OH to the other side.
In case hydrogen is still unbalanced, then balance by adding one OH-, for every
excess of H atom on the same side and one H2O on the other side.
(c) Equalize the charge on both sides by adding suitable number of electrons to the side
deficient in negative charge.
Multiply the two half reactions by suitable integers so that the total number of electrons
gained in one half reaction is equal to the number of electrons lost in the other half
reaction.
Add the two balanced half equations and cancel any term common to both sides.
H
Illustration 3. (a) NO3 H2 S
HSO4 NH4
acidic medium
OH
(b) Fe N2H4
Fe(OH)2 NH3
alkaline medium
Solution:
(a) Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
N5+ + 8e N3S 2- S 6+ + 8e
N5+ + S2- N3- + S 6+
NO-3 + H2 S NH+4 + HSO-4
No other atom (except H and O) is unbalanced therefore no
need for this step.
Balancing O atom. This is made by using H2 O and H+ ions. Add
desired molecules of H2O on the side deficient in O atom and
double H+ on opposite side. Therefore
NO-3 + H2 S + H2O NH+4 + HSO 4- + 2H+
Step 6:
(b) Step 1:
Balance charge by H+
NO-3 + H2 S + H2O + 3H+ NH+4 + HSO-4 + 2H+
Balanced equation is
NO-3 + H2 S + H2 O + H+ NH+4 + HSO4Fe
Fe2+ + 2e
3N22 + 2e 2N
Step 2:
Step 3:
Fe + N22
Fe + N2H4
Fe2+ + 2N3-
Fe(OH)2 + 2NH3
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Step 4:
Step 5:
Step 6:
No other atom (except H and O) is unbalanced and therefore no
need for this step.
Fe + N2H4 + 4OH- Fe(OH)2 + 2NH3 + 2H2O
Balance charge by H+
Fe + N2H4 + 4OH- + 4H+ Fe(OH)2 + 2NH3 + 2H2 O
Finally, Fe + N2H4 + 2H2O Fe(OH)2 + 2NH3
Exercise 4.
(i) Write the balanced equation when ferrous sulphate is treated with acidified
(H2SO4) potassium permanganate.
(ii) Balance the equation MnO - + C O 2 - + H + → CO + Mn 2 + + H O
4
2 4
2
2
Stoichiometry is related to the number of atoms or molecules of reacting species and product
formed during the course of reaction which is governed by the law of chemical combination
(discussed previously in Some Basic Concepts).
Here, we focus our attention about the stoichiometry related to redox reaction and acid and base
neutralization reaction. As stoichiometric calculations are necessary for both analytical purpose
(to find out chemical composition of given sample) and for assesing required amount of raw
material to obtain desired mass of products.
For analytical purpose, most reactions held in liquid state, (i.e. aqueous solution) as solution
provides better homogenity which helps in accurate measurement with minimum error. This type
of analysis in aqueous solution is known as volumetric anlalysis.
VOLUMETRIC ANALYSIS
Important definitions:
No. of moles of solute
(I)
Molarity =
Vol. of solution in L
(II)
Normality =
No. of gram equivalents of solute
Vol. of solution in L
No. of gram equivalents of solute =
Weight of solute
Equivalent weight
Molecular weight
'n'
It can be inferred that number of gram equivalents of a substance = n number of moles and
Normality= ‘n’ Molarity
It is extremely convenient to use the law of gram equivalents to solve problems based on
chemical reactions. According to this law the ‘number of gram equivalents of all reactants are
equal to each other in a reaction assuming none of them are in excess and is also equal to
number of gram equivalents of all products’ assuming that all the reactants are undergoing
reaction in the reaction.
Equivalent weight =
We can use this law conveniently to solve problems without requiring to know much about the
reactions. For this we need to have good understanding of the ‘n’ factor of a substance. ‘n’ factor
is the valency factor or conversion factor.
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
9
*
*
(1)
(2)
n – factor of substance in redox reaction is equal to number of moles of lost or gain
electron per molecule.
n – factor of substance in non – redox reaction is equal to the product of displaced mole
and its charge.
+
Acids: The number of moles of replaceable H ions per mole of the acid e.g. for
HCl n = 1; H2SO4 n = 2; H3PO4 n = 3
Bases: The number of moles of replaceable OH- ions per mole of the base e.g.
NaOH n = 1; Ba(OH)2 n = 2; Al(OH)3 n = 3
Illustration 4. H3PO4 is a dibasic acid and one of its salts is NaH2PO4. What volume of 1 M
NaOH solution should be added to 12 g of NaH2PO4 to convert it into Na3PO4?
(A) 100 ml
(B) 200 ml
(C) 80 ml
(D) 300 ml
Solution.
(B). Eq. of NaH2PO4 = Eq. of NaOH
12
2 1 V 10 3
M
V 200 ml
Illustration 5. A sample of H2SO4 (density 1.787 g ml–1) is labelled as 86% by weight. What is
molarity of acid? What volume of acid has to be used to make 1 litre of 0.2 M
H2SO4?
Solution:
Let us take 1 lit of the solution
Weight of the solution = 1000 1.787 g = 1787 g
86
Weight of H2SO4 1787
gm
100
1787 86
Number of moles of H2SO4 =
= 15.68
100 98
Hence molarity of solution of H2SO4 is = 15.68
Let V ml of this H2SO4 are used to prepare 1 litre of 9.2 M H2SO4
milli mole of conc. H2SO4 = milli mole of dil. H2SO4
V 15.68 = 1000 0.2 V = 12.75 ml
Illustration 6. Calculate the amount of NaOH required to neutralize 25 meq. of the following
(a) HCl
(b) N2O5
Solution:
25 meq. of each separately react with NaOH and therefore only 25 meq. of
NaOH are required every time.
Meq. of NaOH required = 25 or
W
1000 = 25
40
W=1g
Exercise 5.
Calculate the amount of oxalic acid (H2C2O4.2H2O) required to obtain 250 ml of
deci–molar solution.
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
10
)
(
2
)
2
(
)
(
2
)
(
)
3
(
Exercise 6.
Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the
reaction:
CaCO s + 2HCl aq. CaCl aq. + CO g + H O
What mass of CaCO3 is required to react completely with 25 ml of 0.75 M HCl?
Exercise 7.
0.968 g of an acid are present in 300 ml of a solution. 10 ml of this solution required
exactly 20 ml of 0.05 N KOH solution. Calculate no. of neutralisable protons and
equivalent weight of the acid (Given mol. wt. of acid is 98)
Exercise 8.
A 6.90 M solution of aqueous KOH has 30% by weight of KOH. Calculate density of
solution.
(3)
Salts
Salts which reacts in such way that no atom in the salt undergoes change in oxidation
state (oxidation state of an element in a molecule is the charge the element would have if
all the bonds associated with the elements are assumed to be completely ionic.)
‘n’ factor = number of moles of cation in one mole of the salt oxidation state of the
cation.
‘n’ factor of MgCl2 = 1 2 = 2
(Note: whenever two substances react such that their ‘n’ factors are in the ratio of x : y,
the molar ratio of these substances in the balanced chemical reaction would be in the
ratio of y : x)
Examples
KMnO 4
(i)
7
(ii)
n=5
KMnO 4
acidic
Mn2
2
(iv)
(v)
n=3
KMnO 4
7
n=1
K 2Cr2 O7
12
2
(vi)
basic
Mn6
0
(ix)
6
acidic
Cr 3
(x)
(xi)
3
FeSO 4
S 2 O32
4
6
n =1
Fe2O3
I2
(viii)
4
n=6
FeSO 4 Fe2 O3
C2 O42
+6
neutral
Mn4
7
(iii)
(vii)
(xii)
n=1
CuS
2
CO2
+8
n=2
I
2
n=2
S 4 O62
5
Cu2
SO 2
4
n=6
IO3 ICl
+5
+1
n=4
SO 2 SO24
4
6
+6
+4
n=2
n=2
‘n’ factor for a redox reaction is the moles of electrons released or acquired by 1 mole of
the reactant in the reaction.
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
11
Illustration 7. Calculate the mass of anhydrous Ca3(PO4)2 present in 250 ml of 0.25 M solution.
Solution:
Meq. of Ca3(PO4)2 = 250 0.25 6 = 375
W
1000 = 375
51.66
W = 19.3725 g
Illustration 8. If 1.58g of KMnO 4 in acidic medium completely reacts with ferrous oxalate
(FeC2 O4 ), what weight (in g) of ferrous oxalate will be required?
(A) 2.73
(C) 11.19
Solution:
(B) 4.73
(D) 8.5
(A). nfactor of ferrous oxalate = 3
nfactor of KMnO 4 in acidic medium = 5
1.58
102
158
No. of eq. of KMnO 4 5 102
No. of moles of KMnO 4 =
No. of eq. of ferrous oxalate = 5 102
5 102
Mass of ferrous oxalate =
144 g = 2.40 g
3
Illustration 9. In the reaction, VO Fe2O3 FeO V2 O5 , the equivalent weight of V2O5 is
equal to its
Solution:
(A) Mol. weight
(B) Mol. weight
8
(C) Mol. weight
6
(C). n–factor for V2O5 = 2 (5 – 2) = 6
M
E
6
(D) None of these
Illustration 10. Calculate the equivalent weights of each oxidant and reductant in
FeSO4 + KClO3 KCl + Fe2(SO4)3
Solution:
Eq. wt. of oxidant or reductant =
Mol. wt. of oxidant or reductant
No. of moles of 'e' lost or gained by the substrate
Fe2+ Fe3 e
Eq. wt. of FeSO4 =
Cl
+5
+ 6e Cl
Mol. wt. of FeSO 4
152
=
= 152
1
1
–1
Eq. wt. of KClO3 =
Mol. wt. of KClO3
122.5
=
= 20.42
6
6
Illustration 11. What is the weight of sodium bromate and molarity of solution to prepare 500 ml
of 0.6 N solution when half cell reaction is
+
–
BrO3- + 6H + 6e Br + 3H2O
Solution:
Meq. of sodium bromate = 500 0.6 = 300
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12
W
1000 = 300
E
W
M
or
1000 = 300 ENaBrO3
151
6
6
300 151
0.6
W=
= 7.55g and molarity =
= 0.1 M
6000
6
Illustration 12. Calculate the equivalent weights of each oxidant and reductant in:
(a) Na SO + Na CrO Na SO + Cr OH
2
3
2
4
2
4
3
(b) KI + K Cr O Cr 3 + + 3I
2
2 7
2
Solution:
M 126
63
2
2
M 162
Equivalent weight of Na2CrO4 =
54
3
3
M 166
(b) Equivalent weight of KI =
166
1
1
M 294
Equivalent weight of K2Cr2O7 =
49
6
6
(a) Equivalent weight of Na2SO3 =
Exercise 9.
Find the amount of 98% pure Na2CO3 required to prepare 5 litres of 1 N solution.
Exercise 10.
What is the equivalent weight of potash alum (K2SO4 Al2(SO4)3.24H2O)?
Exercise 11.
What is equivalent weight of FeC2O4 in the following reaction?
+3
FeC2O4 Fe + CO2
Titration: The procedure for determining the concentration of a solution by adding its known
volume to react completely with other solution of known strength whose volume is justified by
experiment.
Solution of known strength is called standard solution and are classified as
(i) Primary standard
(ii) Secondary standard
Substances preferred primary standards must have following characteristics:
(i) Easily available and easy to preserve (made stable) {resistant to moisture and air)
(ii) Readily soluble in given solvent.
(iii) The reaction with a standard solution should be of constant stoichiometry.
(iv) Titration error should be negligible even at moderate concentration.
Some common examples which can be used as primary standard
Potassium hydrogen phthalate C8H5 O4K 204.23
-
Acid base
Anhydrous sodium carbonate Na2 CO3 106
-
Acid base
Potassium dichromate [294.19]
Arsenic oxide As2O3 [197.85]
-
Redox
Redox
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
13
Potassium iodate KIO3 [214]
Sodium oxalate Na2C2O4 [134]
EDTA [Na] [372.3]
-
Redox
Redox
Complexo method
The aim of titration is the addition of a quantity of standard and solution chemically equivalent to
the quantity of unknown.
The completion of reaction is often indicated by a change in colour of reaction mixture which may
be either due to change in colour of reactant or a substance mixed externally known as indicator.
TYPE OF TITRATIONS
(i) Acid – base or neutralization titration: The reaction in which an acid reacts with a base to
give salt and water known as neutralization reaction and titration involving such a reaction
known as neutralization titration.
(ii) Redox titration: Titration involving oxidation-reduction known as redox – titration.
(iii) Precipitation titration: When two such solution are mixed during the course of titration a
precipitate is formed and the end point is indicated by completion of precipitation.
Titration involving such type is known as precipitation titration. Some common examples are
KCl AgNO3 AgCl KNO3
White ppt.
KI AgNO3
AgI KNO3
ppt.
Some time titration involving silver nitrate also known as argentometric titration as all other
salt of silver are insoluble in water thus form ppt.
To find out the concentration or compositon of given samples, titration is used in different mode
named as simple titration, double titration, back titration as discussed below.
SIMPLE TITRATION
In this, we can find the concentration of a substance with the help of the conc. of another
substance which can react with it.
For example: Let there be a solution of a substance A of unknown concentration. We are given
another substance B whose concentration is known (N1 ) . We take a certain known volume of A in
a flask (V2 ) and then we add B to A slowly till all the A is consumed by B. (This can be known
with the help of indicators). Let us assume that the volume of B consumed is V1 . According to the
Law of equivalents, the number of gm equivalents of A is equal to the number of gm equivalents
of B.
N1V1 N2 V2 , where N2 is the conc. of A.
From this we can calculate the value of N2 .
llustration 13. In the titration of CH3COOH agaisnt NaOH, we cannot use the
(A) methyl organge
(B) methyl red
(C) phenolphthalein
(D) bromothymol blue
Solution:
(A), (B), (D)
BACK TITRATION
Back titration is used to calculate % purity of a sample. Let us assume that we are given an
impure solid substance C weighing w gms and we are asked to calculate the percentage of pure
C in the sample. We will assume that the impurities are inert. We are provided with two solutions
A and B, where the concentration of B is known (N1 ) and that of A is not known. This type of
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
14
titration will work only if the following conditions are satisfied (a) A, B and C should be such
compounds that A and B can react with each other, A and C can react with each other but
product of A and C should not react with B.
Now we take a certain volume of A in a flask (A taken should be such that gm equivalents of A
gm equivalents of C in the sample. This can be done by taking A in excess). Now we perform a
simple titration using B. Let us assume that the volume of B used is V1 . In another beaker, we
again take the solution of A in the same volume as taken earlier. Now, C is added to this and after
the reaction is complete, the solution is being titrated with B. Let us assume that volume of B
used up is V2 .
Gram equivalents of B used in the first titration = N1V1
gm. equivalents of A initially = N1V1
gm. equivalents of B used in the second titration is N1V2
gm. equivalents of A left in excess after reacting with C = N1V2
gm. equivalents of A that reacted with C = N1V1 - N1V2
gm. equivalents of pure C = N1V1 - N1V2 .
If the ‘n’ factor of C is x, then the moles of pure C =
N1V1 N1V2
x
N1V1 N1V2
Molecular weight of C.
x
N V N1V2 Molecular wt. of C
percentage of C = 1 1
100
x
w
the weight of C =
Exercise 12.
A sample of chalk weighing 1.5 gms was dissolved in 200 ml 0.1 M dil HCl. The
solution required 50 ml of 0.2 N NaOH to neutralize the excess acid. What is the
weight of CaCO3 in the sample?
DOUBLE TITRATION
The method involves two indicators (Indicators are substances that change their colour when
a reaction is complete) phenolphthalein and methyl orange. This is a titration of specific
compounds. Let us consider a solid mixture of NaOH, Na2CO3 and inert impurities weighing w g.
You are asked to find out the % composition of mixture. You are also given a reagent that can
react with the sample, say, HCl along with its concentration (M1 ).
We first dissolve the mixture in water to make a solution and then we add two indicators in it,
namely phenolphthalein and methyl orange. Now, we titrate this solution with HCl.
NaOH is a strong base while Na2 CO3 is a weak base. So it is safe to assume that NaOH reacts
with HCl first, completely and only then does Na2 CO3 react.
NaOH + HCl NaCl + H2 O
Once NaOH has reacted, it is the turn of Na2CO3 to react. It reacts with HCl in two steps:
Na2 CO3 + HCl NaHCO3 + NaCl and then,
NaHCO3 + HCl NaCl + CO 2 + H2 O
As can be seen, when we go on adding more and more of HCl, the pH of the solution keeps on
falling. When Na2 CO3 is converted to NaHCO3 completely, the solution is weakly basic due to
the presence of NaHCO3 (which is a weaker base as compared to Na2CO3). At this instant
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
15
phenolphthalein changes colour since it requires this weakly basic solution to change its colour.
Therefore, remember that phenolphthalein changes colour only when the weakly basic NaHCO3
is present. As we keep adding HCl, the pH again falls and when all the NaHCO3 reacts to form
NaCl, CO2 and H2 O . The solution becomes weakly acidic due to the presence of the weak
acid H2 CO3 (CO2 + H2 O) . At this instance methyl orange changes colour since it requires this
weakly acidic solution to do so. Therefore, remember methyl orange changes colour only when
H2CO3 is present.
Now, let us assume that the volume of HCl used up for the first and the second reaction,
i.e. NaOH+HCl NaCl+H2 O and Na2 CO3 +HCl NaHCO3 +NaCl be V1 (this is the
volume of HCl from the beginning of titration up to the point when phenolphthalein changes
colour).
Let
the
volume
of
HCl
required
for
the
last
reaction,
i.e.,
NaHCO3 + HCl NaCl+CO2 +H2 O be V2 (this is the volume of HCl from the point where
phenolphthalein had changed colour upto the point when methyl orange changes colour). Then,
moles of HCl used for reacting with NaHCO3 = moles of NaHCO3 reacted = M1V2
moles of NaHCO3 produced by the Na 2CO3 = M1V2
moles of Na2 CO3 that gave M1V2 moles of NaHCO3 = M1V2
Mass of Na 2CO3 = M1V2 106
M1V2 106
100
w
moles of HCl used for the first two reactions = M1V1
% Na2CO3
moles of Na2 CO3 = M1V2
moles of HCl used for reacting with Na2 CO3 = M1V2
moles of HCl used for reacting with only NaOH = M1V1 = M1V2
moles of NaOH = M1V1 = M1V2
Mass of NaOH = M1V1 M1V2 40
% NaOH =
M1V1
M1V2
40
w
100
llustration 14. Boric acid (H3BO3) is
(A) monobasic
(C) tribasic
Solution:
(B) dibasic
(D) aprotic
(A), (D)
Illustration 15. A solution contained Na2CO3 and NaHCO3. 25 mL of this solution required 5 mL
of 0.1 HCl for titration with phenolphthalein as indicator. The titration was
repeated with the same volume of the solution but with methyl orange. 12.5 mL
of 0.1 N HCl was required this tine. Calculate the amount of Na2CO3 and
NaHCO3 in the solution.
Solution:
Neutralization reaction with phenolphthalein is
Na2 CO3 HCl NaHCO3 NaCl
while with methyl orange, the reactions are
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
16
Na2 CO3 HCl NaHCO3 NaCl
Na2 CO3 HCl
NaCl H2 O CO2
produced
and
NaHCO3 HCl NaCl H2 O CO 2
originally present
Thus,
We have with phenolphthalein,
m.e. of Na2CO3 = m.e. of 5 mL of 0.1 N HCl
= 0.1 5 = 0.5
eq. of Na2CO3 =
0.5
0.0005
1000
wt. of Na2CO3 = (0.0005 106)g = 0.053g
[Eq. wt. of Na2CO3 in the given reaction in 106]
and with methyl orange,
m.e. of Na2 CO3 m.e. of NaHCO3 m.e. of NaHCO3
produced
originally present
= m.e. pf 12.5 mL of 0.1 N HCl
or 0.5 + 0.5 + m.e. of NaHCO3 = 0.1 12.5 = 1.25
or m.e. of NaHCO3 = 0.25
weight of NaHCO3 =
0.25
84 0.021g
1000
(eq. wt. of NaHCO3 = 84)
Exercise 13.
A solution contains Na2CO3 and NaHCO3. 10 mL of the solution requires 2.5 mL
of 0.1 M H2SO4 for neutralization using phenolphthalein as an indicator. Methyl
orange is added when a further 2.5 mL of 0.2 M H2SO4 was required. Calculate
the amount of Na2CO3 and NaHCO3 in one litre of the solution.
IODOMETRIC AND IODIMETRIC TITRATION
The reduction of free iodine to iodide ions and oxidation of iodide ions to free iodine occurs in
these titration
I2 + 2e- 2I- (reduction)
2I- I2 + 2e- (oxidation)
These are divided into two types
Iodometric titration: In iodometric titrations, an oxidising agent is allowed to react in
neutral medium or in acidic medium with excess of potassium iodide to librate free iodine.
KI + oxidising agent I2
Free iodine is titrated against a standard reducing agent usually with sodium
thiosulphate, Halogen, dichromates, cupric ion, peroxides, etc can be estimated by this
method.
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
17
I2 + 2Na2 S2 O3 2NaI + Na2 S4 O6
2CuSO4 + 4KI Cu2I2 + 2K 2SO 4 + I2
K 2Cr2 O7 + 6KI + 7H2 SO4 Cr2 (SO 4 )3 + 4K 2 SO 4 + 7H2 O + 3I2
Iodimetric titration: These are the titrations in which free iodine is used as it is difficult to
prepare the solution of iodine (volatile and less soluble in water) it is dissolved in KI
solution.
KI + I2 KI3 (Potassium triiodide)
This solution is first standardised before use. With the standard solution of I2, substance
such as sulphite, thiosulphate, arsenite etc. are estimated.
In the iodimetric and iodometric titrations, starch solution is used as indicator. Starch
solution gives blue or violet colour with free iodine. At the end point the blue or violet
colour disappears when iodine is completely changed to iodide.
Potassium permanganate titration: In these titrations, reducing agents like
FeSO 4 ,Mohr 's salt,(NH4 )2 SO 4 .FeSO4 .6H2 O,H2O 2 ,As2O3 ,oxalic acid (COOH)2 and
oxalates
(COONa)2 etc. are directly titrated against KMnO 4 as oxidising agent in acidic medium. For
Example 5Fe2 MnO4 8H 5Fe3 Mn2 4H2 O
Ferrous ion Permagnate ion Ferric ion
COO
5
+ 2MnO 16H 2Mn2 10CO 8H O
4
2
2
COO
Oxalate ion
Potassium dichromate titrations
In these titrations, the above listed reducing agents are directly titrated against K 2 Cr2 O7 as
oxidising agent in acidic medium for example,
6Fe2 Cr2O27 14H 2Cr 3 6Fe3 7H2 O
Ceric sulphate titrations
In these titrations, the reducing agents such as Fe2 salts, Cu salts, nitrites, arsenites, oxalates
etc. are directly titrated against ceric sulphate, Ce(SO 4 )2 as the oxidising agent. For example
2Ce 4 H2 O AsO 43
2Ce3 2H
AsO3
arsenite ion ceric ion
Arsenate ion
Cerous ion
(COOH)2
2Ce 4 2CO2 2Ce3 2H
Oxalic acid
VOLUME STRENGTH OF H2O2
x volumes of H2 O2 means x litre of O 2 is liberated by 1 litre of H2 O2 on decomposition
2H2 O2 2H2 O + O2
68 gm
22. 4 lit at STP
22.4 lit (at STP) of O 2 is given by 68 gm of H2 O2
x litre of O 2 is released from
68x
17x
gm of H2 O2 =
22.4
5.6
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18
x lit of O 2 is given by =
Strength, S =
68x 17x
(strength)
22.4 5.6
17x
5.6
S
17x
x
=
=
[ x = Nx5.6]
E
5.6 34 / 2
5.6
1
x
Molarity =
Normality
2
11.2
Normality =
Illustration 16. 280 ml of H2 O2 reacts completely with 15.8 g of KMnO 4 in acidic medium, then
the volume strength of H2 O2 is
(A) 10
(C) 5
Solution:
(B) 100
(D) 20
(A). Let the volume strength of H2 O2 be x
According to the question
meq. of H2 O2 meq. of KMnO 4
x
15.8
1000
5.6 158 / 5
15.8 5
50x
1000
158
50 x = 500 x 10
280
Illustraton 17. 0.2 g of a sample of H2O2 required 10 ml of 1 N KMnO4 in a titration in the
presence of H2SO4. Purity of H2O2 is
(A) 25%
(B) 85%
(C) 65%
(D) 95%
Solution:
(B). meq. of H2O2 = 10
Weight of H2O2 10 103 17
0.17
% purity of H2O2
100 85%
0.20
Illustration 18. A polyvalent metal weighing 0.1 gm and having atomic weight of 51 reacted with
dilute H2SO4 to give 43.90 ml of hydrogen at N.T.P. This solution containing the
metal in the lower oxidation state was found to require 58.8 ml of 0.02 M KMnO4
for complete oxidation. What are the oxidation states of the metal in the two
reactions?
Solution:
Let lower oxidation state of the metal be n
0.1 n
43.90
Equivalents of metal =
= equivalents of H2 evolved =
2
51
22400
n = 2
Let final oxidation state = n
Then equivalents of metal = equivalents of oxidant
0.1 n 2 58.8 0.02 5
n = 5
51
1000
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
19
Illustration 19. 1.20 gm of sample of Na2 CO3 and K 2 CO3 was dissolved in water to form 100 ml
of a solution. 20 ml of this solution required 40 ml of 0.1 N HCl for complete
neutralisation. Calculate the weight of Na2 CO3 in mixture. If another 20 ml of this
solution is treated with excess of BaCl2 what will be the weight of the precipitate?
Solution:
Let weight of Na2 CO3 = x gm
Weight of K 2 CO3 = y gm
x + y = 1.20 gm
…(1)
For neutralization reaction of 100 ml of
Meq. of Na2 CO3 + Meq. of K 2 CO3 = Meq. of HCl
x
y
40 0.1 100
2 1000
2 1000
106
138
20
69x + 53y = 73.14
…(2)
From equation (1) and (2)
x = 0.5962 gm, y = 0.604 gm
Solution of Na2 CO3 and K 2 CO3 gives ppt. of BaCO3 with BaCl2
(meq. of Na2 CO3 + Meq. of K 2 CO3 ) in 20 ml = meq. of BaCO3
Meq. of HCl for 20 ml mixture = Meq. of BaCO3
Meq. of BaCO3 = 40 0.1 = 4
WBaCO3
1000 = 40 0.1 = 4
MBaCO3
WBaCO3
2 1000 = 4
197
WBaCO3 = 0.394 gm
Illustration 20. What is the weight of sodium bromate and molarity of solution to prepare 85.5
mL of 0.672 N solution when half cell reaction is
BrO3– + 6H+ + 6e Br– + 3H2O
Solution:
Meq. of sodium bromate = 85.5 0.672 = 57.456
Let weight of sodium bromate be w
M
Eq. wt. of sodium bromate =
(as n-factor = 6)
6
W
1000 = 57.456, WNaBrO3 = 1.446 gm
151/ 6
As molarity n-factor = Normality
0.672
0.672
Molarity =
= 0.112M
n factor
6
Illustration 21. A solution 0.1 M KMnO4 is used for the reaction
S2 O32- + 2MnO4- +H2 O MnO 2 + SO 42- + OH-
What volume of solution in mL will be required to react with 0.158gm of Na2 S2 O3
Solution:
Redox changes are: Mn7+ + 3e Mn+4
S2+
2S6+ + 8e
2
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
20
Meq. of KMnO4 = Meq. of S2O32Let volume of KMnO 4 required = V ml
0.158
8 1000 ( MNa2 S2 O3 = 158, n-factor = 8)
158
V = 26.67 ml
n- factor for MnO -4 = 3
V 0.1 3 =
Illustration 23. 25 ml of H2 O2 solution were added to excess of acidified solution of KI. The
iodine so liberated required 20 ml of 0.1 N Na2 S2O3 for titration. Calculate the
strength of H2 O2 in terms of normality and volume strength.
Solution:
Redox change is
O -12 + 2e 2O 22I- I2 + 2e
2S2+
S+5/2
+ 2e
2
4
I2 + 2e 2I-
Meq. of H2 O2 = Meq. of I2 = Meq. of Na2S2O3 = 20 0.1=2
NH2 O2
Meq.
Volume in
ml
2
25
Volume strength (x) = N 5.6
2
Volume strength =
5.6=0.448 volume
25
Exercise 14.
0.2M KMnO4 solution completely reacts with 0.05 M FeSO4 solution under acidic
conditions. The volume of FeSO4 used is 50 ml. What volume of KMnO4 was used?
Exercise 15.
10 g. of a sample of Ba(OH)2 is dissolved in 10 ml. of 0.5N HCl solution. The excess of
HCl was titrated with 0.2N NaOH. The volume of NaOH used was 20 cc. Calculate the
percentage of Ba(OH)2 in the sample.
Exercise 16.
0.2 gm of a solution of mixture of NaOH and Na2CO3 and inert impurities was first
titrated with phenolphthalein and N/10 HCl 17.5 ml of HCl was required at the end
point. After this methyl organge was added and 2.5 ml of same HCl was again required
for next end point. Find out percentage of NaOH and Na2CO3 in the mixture.
Exercise 17.
A solution of H2 O2 , labelled as ‘20 volumes’, was left open. Due to this some H2 O2 ,
decomposed and the volume strength of the solution decreased. To determine the new
volume strength of the H2 O2 , solution, 10 mL of the solution was taken and it was
diluted to 100 mL. 10mL of this diluted solution was titrated against 25 mL of 0.0245 M
KMnO4 solution under acidic condition. Calculate the volume strength of the
H2 O2 solution.
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
21
HARDNESS OF WATER
Hard water with soap forms insoluble precipitates of calcium and magnesium salts of fatty acids.
2RCOONa Ca / Mg
(RCOO)2 Ca / Mg 2Na
Hardness of water is of two types:Temporary hardness and Permanent hardness:
(i)
Temporary Hardness: It is due to the presence of bicarbonates of calcium and
magnesium in water.
Process of removal of temporary hardness:
(a)
By boiling of water:
M(HCO3 )2
MCO3 H2 O CO2
(b)
Clarke’s process: By the addition of Ca(OH)2
Ca(HCO3 )2 Ca(OH)2
2CaCO3 2H2 O
Mg(HCO3 )2 Ca(OH)2 MgCO3 CaCO3 2H2 O
(c)
By the addition of Na2CO3:
Ca(HCO3 )2 Na 2CO3 CaCO3 2NaHCO3
(ii)
Permanent hardness: It is due the presence of chlorides and sulphates of Ca2+ and
Mg2+.
Removal of permanent hardness:
(i)
(ii)
By adding Na2CO3 or Na3PO4:
CaSO4 Na 2CO3 CaCO3 Na2 SO 4
3CaCl2 2Na3PO4 Ca3 (PO 4 )2 6NaCl
Permuit process:
Ca2 Na2 Z CaZ 2Na
Mg2 Na2 Z MgZ 2Na
(iii)
Na2Z is the sodium
zeolite (Na2Al2Si2O8.xH2O)
Calgon (calcium gone) process:
Na6 (PO3 )6 is called calgon which is written as Na2 Na4 (PO3 )6 .
2
2
M2
Na2M(PO3 )6 2Na [where M Ca / Mg ]
(iv)
Ion exchange resins process:
2R H Ca2
(Rn )2 Ca 2H
Acidic resin n
cation exchanger
Degree of hardness: Degree of hardness is defined as number of parts by weight of CaCO3 (or
its equivalent quantities of other substance) present in million parts by weight of water.
Weight of CaCO3
Hardness of water
106 ppm
Weight
of
water
Na 4 (PO3 )6
N
Na2CO3 reagent. After boiling
50
the volume was again made to 200 ml and the solution filtered. 25 ml of the
N
filtrate required 8.2 ml of
HCl for neutralization. Calculate the permanent
50
hardness in ppm.
Illustration 24. 200 ml of hard water was boiled with 100 ml
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
22
Solution:
Total Na2CO3 added in sample
100 1
2 Meq.
50
8.2 1
Meq
50
Total Meq. of HCl required for 200 ml filtrate = Meq. of Na2CO3 in sample
8 8.2 1
1.312
50
Used Na2CO3 = 2 – 1.312 = 0.688 Meq. = Meq. Of CaCO3
Weight of CaCO3 0.688 50 10 3 3.4 102 gm
Number of Meq of HCl reacted with 25 ml filtrate
Hardness of water
Weight of CaCO3
3.4 10 2
10 6
106 172 ppm
Weight of water
200
Exercise 18.
A sample of drinking water was found to be severely contaminated with chloroform
(CHCl3), supposed to be carcinogen. The level of contamination was 15 ppm (by
mass)
(i)
Express this in % by mass.
(ii) Determine the molarity of chloroform in the water sample.
% OF FREE SO3 IN OLEUM
x % of labelled oleum (mixture of H2 SO 4 SO3 ) represents that x gm of H2SO4 formed when
100 g of oleum diluted with water. Therefore weight of dilute water = (x – 100) g
x 100
Moles of free SO3 = Moles of water (required for dilution)
18
Illustration 25. Calculate the % of free SO3 in an oleum, that is labelled 118% H2SO4.
Solution:
Oleum is mixture of H2 SO 4 SO3 H2 S2 O7
If initial weight of labelled H2S2O7 (oleum) = 100 gm
Then weight of H2SO4, after dilution = 118 gm
Weight of H2O = 18 gm
or Moles of H2O = Moles of SO3 18 1
18
Weight of SO3 1 80 gm
% of SO3
Weight of SO3 100
80%
100
SOME IMPORTANT REACTIONS REGARDING STOICHIOMETRY
(i)
Effect of Heat on Carbonate and Bicarbonate
(a) (i) Na2 CO3
no effect
(ii) K 2CO3
no effect (as these are thermally stable)
(b) NH4 2 CO3
2NH3 CO2 H2 O
(c) Other carbonate decompose to gives oxides and CO 2
(i) MCO3
MO CO2
(ii) Ag2 CO3
2Ag CO2
Strong
1
O2
2
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
23
(d)
All bicarbonates decomposes to carbonates, CO 2 and H2O
(i)
2NaHCO3
Na2CO3 CO2 H2 O
Solid
(ii) Mg HCO3 2
MgCO3 CO2 H2 O
Solution
(ii)
Effect of Heat on Chlorides, Bromides and Iodides
(a)
Generally these are not effected by heat.
(b)
Some halides of higher oxidation states changes into halides of lower oxidation
state on heating.
3
2
2
1
2FeCl3
2FeCl2 Cl2 (ii) 2CuCl2
2CuCl Cl2
(c)
Hydrated halides on heating convert to oxides, H2O and haloacids (HF, HCl, HBr,
HI) such as
(i) MgCl2 .6H2O
MgO 2HCl 5H2 O
(ii) 2AlCl3 .6H2O
Al2 O3 6HCl 5H2O
White
(iii) 2CrCl3 .6H2 O
Cr2O3 6HCl 9H2 O
(d)
NH4 Cl s
NH3 HCl
sublination
(iii)
Effect of heat on Sulphites and Bisulphites
(a) 4Na2 .SO3
3Na2 SO4 Na2 S (b) 2NaHSO3
Na2 SO3 SO2 H2 O
(iv)
Effect of Heat on Sulphates, Pyrosulphates and Bisulphates
Sodium and potassium sulphates are thermally stable. However other sulphates,
anhydrous or hydrous, decompose to give different products. Such as
(a)
2 NH4 2 .SO 4
N2 6H2 O 3SO2 2NH3
(b)
(c)
1
CaSO4
CaO SO2 O2
2
Fe2 SO4 3
Fe2 O3 3SO3
Yellow
(d)
2FeSO4
Fe 2O3 SO3 SO2
light green
(e)
Blackish brown
Re d
CuSO4 .H 2 O
CuO SO3 H2O
Blue virtriol
(f)
Na2 S2 O7
Na2 SO4 SO3
Sodium
pyrosulphate
(g)
(v)
2NaHSO4
Na2 SO 4 H2 O SO3
Effect of heat on Nitrates
They undergo thermal decomposition in following ways
(a)
NH4NO3
N2 O 2H2O
1
O2
2
(b) (i)
NaNO3
NaNO2
(c) (i)
1
450 C
AgNO3
AgNO2 O 2
2
1
(ii) KNO3
KNO 2 O2
2
1
(ii) AgNO3
Ag NO2 O2
2
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
24
1
(iii) HgNO3
Hg NO2 O2
2
(d)
1
(i) Mg NO3 2
MgO 2NO2 O2 ,
2
1
(ii) Pb NO3 2
PbO 2NO2 O2
yellow
2
white
1
(iii) Cu NO3 2
CuO 2NO 2 O2
2
(vi)
Effect of heat on Oxide and Hydroxide
Hydroxide, and higher oxides and some metals undergo following change on heating
strongly.
(i) 2Al OH3
Al2 O3 3H2 O
4
2
(ii) PbO 2
PbO
White
Brown
yellow
1
O2
2
SOME OTHER IMPORTANT REACTIONS
(i)
2CuSO4 4KI Cu2I2 I2 2K 2 SO4
(ii) O3 2KI H2 O
2KOH I2 O2
(iii) 2MnO4 5C2 O24 16H 2Mn2 10CO2 8H2 O
(iv) 2KMnO4 10FeSO 4 8H2SO 4 2MnSO 4 5Fe2 SO4 3 8H2O K 2SO 4
(v) BaCO3 2HCl
BaCl2 CO 2 H2O
(vi) BaCl2 H2CrO 4 BaCrO 4 2HCl
(vii) 2BaCrO 4 6KI 8H2 SO 4 3I2 2BaSO4 3K 2 SO4 Cr2 SO 4 3 8H2 O
(viii) Mn3 O4 2FeSO 4 4H2 SO 4 3MnSO 4 Fe2 SO4 3 4H2 O
3
O2
2
SOME IMPORTANT POINTS TO REMEMBER
(ix) KClO3
KCl
(i)
(ii)
If a salt of strong acid and strong base (e.g. NaCl) is present in the solution it will not
react with any acid or base added to the solution.
If a salt of strong acid and weak base (e.g. NH4 Cl ) is present in the solution it will only
react with strong base but it will not react with strong acid.
NH4 Cl NaOH NaCl NH4 OH
(iii)
If a salt of strong base and weak acid e.g. CH3 COONa is present in the solution it will
react only with strong acid but it will not react with strong base.
CH3 COONa HCl
NaCl CH3 COOH
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
25
ANSWER TO EXERCISES
Exercise 1:
(i)
Al
Al3 (aq)3e (Oxidation half reaction)
3Ag+ 3e
(ii)
2Al
3Ag
(Re duction half reaction)
3
2Al 6e(Oxidation half reaction)
2+
3Cu (aq)+6e- 3Cu(Re duction half reaction)
Exercise2:
Oxidants: (i)H2O 2
(ii)NaClO3
Reductants (i)[Fe(CN)6 ]4
(ii)I2
Exercise 3:
(i) + 2.5, (ii) +7 (iii) +6 (iv) 0
Exercise 4:
(i) Step (i)
KMnO 4 +H2 SO 4 +FeSO 4 K 2 SO 4 + MnSO 4 + Fe2 (SO 4 )3 + H2O
7
2
3
2
Step (ii)
KMnO 4 H2 SO4 FeSO 4 Κ 2 SO4 MnSO 4 + Fe 2 (SO 4 )3 + H2O
is reduced to Mn2+ (+ 7 +2) and Fe+2 is oxidized to
Thus here Mn+7
Fe+3 (+2 +3).
Thus
Oxidation:
2Fe+2
Fe+3
2 2
Reduction: Mn+7 Mn+2 5
Step (iii)
2Fe+2 Fe+3
2 + 2e
Mn+7 + 5e- Mn+2
Step (iv)
Multiplying equation (a) by 5 and (b) by 2.
10Fe+2 5Fe+3
2 + 10e
2Mn+7 + 10e- 2Mn+2
+2
Adding, 10Fe+3 +2Mn+7 5Fe+3
2 + 2Mn + H2O
Thus the required equation may be written as
10FeSO 4 +2KMnO 4 +H2 SO 4 5Fe2 (SO 4 )3 +2MnSO4 +K 2 SO 4 + H2 O
Step (vi)
to balance SO 42 ions; multiply H2 SO4 by 8
10FeSO 4 +2KMnO4 +8H2SO 4 5Fe2 (SO 4 )3 + 2MnSO4 + K 2SO 4 + H2 O
2
7
3
4
(ii) MnO 4 + C2 O 42 + H+ CO2 + Mn2 +H2 O
Thus here Mn+2 is reduced to Mn+2 (+7 +2) and carbon in C2 O42
is oxidised to
CO 2
Thus here Mn+2 is reduced to Mn+2 (+7 +2) and carbon in C2 O-24 is oxidised
to CO 2 . Thus,
3
Oxidation:
4
C2 O42 2CO 2 2e 2
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
26
Reduction:
Mn+7 + 5e- Mn+2 5
3
5C2 O42 10CO2 10e
2Mn+7 + 10e - 2Mn+2
3
adding5C2 2Mn2 10CO2 Mn2
2MnO -24 5C2 O-24 2Mn+2 +10CO2
+
Making provision of H on the L.H.S since these are given in the required reaction. It
+
must be 16H because R.H.S has 8H2O. Thus the balanced equation is
2MnO 4 + 5C2 O-24 16H 2Mn2 10CO2 8H2 O
Exercise 5:
Molarity of solution
1
M
10
Volume of solution = 250 ml
Millimole of oxalic acid M V(ml)
1
250
10
W
1
1000
250 [Molecular weight of oxalic acid = 126]
126
10
250 126
W
3.15 g
10 1000
Exercise 6:
CaCO3 (s) 2HCl(aq.) CaCl2 (aq.) CO2 (g) H2O( )
meq. of CaCO3 = meq. of HCl
W
1000 25 0.75 1
100
2
WCaCO3 0.9375 g
0.94 g
Exercise 7:
Number of neturalizable protons = 3 (H3PO4), Equivalent weight = 32.268 gram.
Exercise 8:
1.288 gram/ml
Exercise 9:
270.408 gram
Exercise 10:
118.5 gram
Exercise 11:
48 g
Exercise 12:
33.33%
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
27
Exercise 13:
The neutralization reactions are:
2Na2 CO3 H2 SO4
2NaHCO3 Na 2 SO 4
2NaHCO3 H2 SO 4 Na2 SO4 H2 O CO 2
The volume of H2SO4 (2.5 mL) used while using phenolphthalein corresponds to
the volume required for conversion of Na2CO3 to NaHCO3 while volume of H2SO4
(2.5 mL) further added corresponds to the volume required for conversion of
NaHCO3 to Na2SO4. Thus at the end point with phenolphthalein, we have,
m.e. of 2.5 mL of 0.1 M (i.e. 0.2 N) H2SO4) = m.e. of Na2CO3
or 2.5 0.2 = m.e. of Na2CO3 or m.e. of Na2CO3 = 0.5
Equivalent of Na2CO3 =
0.5
1000
Weight of Na2CO3/10 mL =
0.5
106 0.053 g
1000
(equivalent weight of Na2CO3 is 106 according to given reaction)
weight of Na2CO3 per litre = 5.3 g
Again with methyl orange, we have,
m.e. of 2.5 mL of 0.2 (i.e., 0.4 N) H2SO4 solution
= m.e. of NaHCO3 produced froom Na2CO3 + m.e. of NaHCO3 originally present.
Since m.e. of NaHCO3 (produced) = m.e. of Na2CO3
2.5 4 = m.e. of Na2CO3 + m.e. of NaHCO3 originally present.
1 = 0.5 + m.e. of NaHCO3 originally present.
m.e. of NaHCO3 originally present = 1 – 0.5 = 0.5
equivalent of NaHCO3
0.5
1000
wt. of NaHCO3 per 10 mL =
0.5
84 0.042 g
1000
(eq. wt. of NaHCO3 = 84 according to given reaction)
Wt. of NaHCO3 per pitre = 4.2
Exercise 14:
0.2 5 V = 0.05 50 V = 2.5 ml.
Exercise 15:
Milli eq. of HCl initially =
10 0.5 = 5
Milli eq. of NaOH consumed = Milli eq. of HCl in excess = 20 0.2 = 4
Milli eq. of HCl consumed = Milli eq. of Ba(OH)2 = 5 4 = 1
eq. of Ba(OH)2
=
1 10-3
Mass of Ba(OH)2
=
1 10-3 (171/2) = 85.5 103 gm
% Ba(OH)2
=
85.5 10 3
100 0.855%
10
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
28
Exercise 16:
Let W1 gm NaOH and W2 gm Na2CO3 was present in mixture. At phenolphthalein end
point.
w1 1 w 2 1
17.5 10 3 -----------
40 2 53 10
(1)
At second end point following reaction takes place
Eq. of NaHCO3 = eq. of HCl used (in second titration) =
1
eq. of Na2 CO3
2
1 W2
1
25
10 3 W2 = 0.0265 gm
2 53
10
Putting the value of W2 in equation (1), we get W1 = 0.06 gm
0.06
100 30%
0.2
0.0265
Percentage of Na2 CO3
100 13.25%
0.2
Percentage of NaOH =
Exercise 17:
Volume strength of a H2 O2 solution: If a solution of H2 O2 is labelled as ‘x volumes’ it
means that 1 mL of the H2 O2 solution on complete decomposition would release
O 2 measuring x mL at STP.
Normality of KMnO 4 solution
=0.0245 5 = 0.1225 N
0.1225 25
3.0625 10 3
1000
Equivalents of H2 O2 in the 10 mL that is titrated = 3.0625 10-3
Equivalents of KMnO 4 used
=
Equivalents of H2O2 in 100 mL=3.0625 10-3 10 = 3.0625 10-2
Equivalents of H2O2 in the original 10 mL = 3.0625 10-2
[ on adding water equivalents of a substance does not change]
3.0625 102
Moles of H2O2 in the original 10 mL =
= 1.53 10-2
2
[‘n’ factor of H2 O2 is 2 because as reacting with KMnO 4 O-1 becomes O -2 ]
1.53 102
1.53 103
10
1.53 103
1
Moles of O 2 that it would give on decomposing =
H2 O2 H2 O O2
2
2
-4
= 7.65 10
Volume of O2 at STP in mL= 7.65 104 22400 = 17.136
Volume strength =17.136
Moles of H2 O2 in 1mL of the original 10 Ml =
Exercise 18:
6
CHCl3 present is 15 ppm or 10 g (or ml) H2O contains 15 g CHCl3.
15
(i)
% by mass
100 1.5 103
6
15
10
(ii)
15
119.5
Molality 6
1.25 10 4 m
10 10 3
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
29
MISCELLANEOUS EXERCISES
Exercise 1:
Calculate the % of free SO3 in oleum (a solution of SO3 in H2SO4) that is labelled
109% H2SO4 by weight?
Exercise 2:
A 5.0 g sample of a natural gas consisting of CH4, C2H4 was burnt in excess of
oxygen yielding 14.5 g CO2 and some H2O as product. What is weight
percentage of CH4 and C2H4 in mixture?
Exercise 3:
A saturated solution is prepared at 70oC containing 32.0 g of CuSO4.5H2O per
100 g solution. 335 g sample of this solution is then cooled to 0oC so that
o
CuSO4.5H2O crystallises out. If the concentration of a saturated solution at 0 C is
12.5 g CuSO4.5H2O per 100 g solution, how much of CuSO4.5H2O is
crystallised?
Exercise 4:
In a gravimetric determination of P, an aqueous solution of dihydrogen
phosphate ion H2PO 4 is treated with a mixture of ammonium and magnesium
ions to precipitate magnesium ammonium phosphate [Mg(NH4)PO4.6H2O]. This
is heated and decomposed magnesium phosphate [Mg2P2O7], which is weighed.
A solution of H2PO4 yielded 1.054 g of Mg2P2O7. What weight of NaH2PO4 was
present originally? (Na = 23, H = 1, P = 31, O = 16, Mg = 24)?
Exercise 5:
What weight of Na2CO3 of 95% purity would be required to neutralize 45.6 ml of
0.235 N acid?
Exercise 6:
A piece of Al weighing 2.7 g is titrated with 75.0 ml of H2SO4 (sp. gr. = 1.18 g/ml
and 24.7% H2SO4 by weight). After the metal is completely dissolved, the
solution is diluted to 400 ml. Calculate molarity of free H2SO4 in solution.
Exercise 7:
To 50 litre of 0.2 N NaOH, 5 litre of 1 N HCl and 15 litres of 0.1 N FeCl3 solutions
are added. What weight of Fe2O3 can be obtained from the precipitate? Also,
report the normality of NaOH left in the resultant solution?
Exercise 8:
Chloride samples are prepared for analysis by using NaCl, KCl and NH4Cl
separately or as mixture. What minimum volume of 5% by weight AgNO3 solution
(sp. gr. = 1.04 g/ml) must be added to a sample of 0.3 g in order to ensure
complete precipitation of chloride in every possible case?
Exercise 9:
25 mL of 0.017 M H2SO3 in strongly acidic solution required the addition of
16.9 mL of 0.01 M MnO4 for its complete oxidation to SO42 or HSO4. In neutral
solution it required 28.6 mL. Assign oxidation number to Mn in each of the
products.
Exercise 10:
50 mL of a solution containing 1 g each of Na2CO3, NaHCO3 and NaOH, was
titrated with N HCl. What will be the titrate reading if
(a) only phenolphthalein is used as indicator?
(b) only methyl orange is used as indicator from the very beginning?
(c) methyl orange is added after the first end point with phenolphthalein?
Exercise 11:
P and Q are two elements which form P2Q3, PQ2 molecules. If 0.15 moles of
P2Q3 and PQ2 weighs 15.9 g and 9.3 g respectively, what are atomic weights of P
and Q?
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
30
Exercise 12:
4 g of an impure sample of CaCO3 on treatment with excess HCl produces 0.88 g
CO2, what is the % purity of CaCO3 sample?
Exercise 13:
Calculate the moles of H2O vapours formed if 1.57 mole of O2 are used in
presence of excess of H2 for the given change.
2H2 O 2
2H2 O
Exercise 14:
Calculate the weight of FeO produced from 2 g VO and 5.75 g of Fe2O3. Also
report the limiting reagent. Given: VO Fe2 O3 FeO V2O5
Exercise 15:
When dissolved in dilute H2SO4, 0.275 g of metal evolved 112 ml of H2 at STP,
calculate equivalent weight of metal.
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
31
ANSWERS TO MISCELLANEOUS EXERCISES
Exercise 1:
109% H2SO4 means that total mass of H2SO4 is 109 g, that would be present
after dilution of 100 g oleum, thus 9 g H2O will combine with all the free SO3 in
100 g of oleum to give a total of 109 g H2SO4.
H2 O SO3 H2 SO4
Moles of free SO3 = Moles of H2O
9
1
18 2
Amount of SO3 = 40 g
Thus, 40 g free SO3 is present in 100 g of the oleum or, % of free SO3 = 40%
Exercise 2:
We know, moles of CO2 formed
CH
4
'a ' mole
14.5
0.330
44
2O2 CO2 2H2O
C2H4 3O 2 2CO2 2H2 O
'b ' mole
CO2 formed = a + 2b = 0.33
Also, a 16 b 28 5
From equation (i) and (ii), we get
a = 0.19 and b = 0.07
WCH4 0.19 6 3.04 g
… (I)
… (ii)
WC2H4 0.07 28 1.96 g
% C2H4
Exercise 3:
1.96
100 39.2% and % CH4 = 100 – 39.2 = 60.8%
5
Weight of CuSO4.5H2O in 335 g solution
32 335
107.20 g
100
Weight of water = 335 – 107.20 = 227.80 g
At 0oC, weight of CuSO4.5H2O in 100 g solution = 12.5 g
Weight of water = 100 – 12.5 = 87.5 g
Thus, during crystallisation amount of H2O remains same. Therefore,
87.5 g H2O has CuSO4.5H2O = 12.5 g
12.5 227.8
227.80 g H2O has CuSO4.5H2O
32.73 g
87
Thus, amount of CuSO4.5H2O crystallised = 107.20 – 32.73 = 74.47 g
Exercise 4:
The given reaction is
NaH2PO 4 Mg NH4 Mg(NH4 )PO 4 .6H2O
Mg2P2 O7
meq. of NaH2PO4 = meq. of Mg2P2O7
W 1000 1.054 1000
120
222
2
4
M
Equivalent weight of NaH2PO4
2
M
Equivalent weight of Mg2P2O7
4
W = 1.1395 g
= weight of NaH2PO4
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
32
Exercise 5:
meq. of Na2CO3 = meq. of H2SO4 (For complete neutralisation)
meq. of Na2CO3 45.6 0.235
W
1000 45.6 0.235
106
2
W = 0.5679
95 g pure Na2CO3 is to be taken then weighed sample = 100 g
100 0.5679
0.5679 g pure Na2CO3 is to be taken, then weighed sample
95
= 0.5978 g
Exercise 6:
Given weight of Al = 2.7 g
2.7
Equivalent of Al
0.3
9
meq. of Al 0.3 1000 300
For H2SO4, given that solution is 24.7% by weight.
Weight of H2SO4 = 24.7 g
and weight of solution = 100 g
100
Volume of solution
ml 84.75ml
1.18
Normality of solution is given by
24.7
NH2 SO4
5.95
100
49
1.18 1000
Now meq. of H2SO4 in 75 ml 5.95 75 446.25 , and
meq. of Al added = 300
meq. of H2SO4 left after reaction = 446.25 – 300
= 146.25
Solution is diluted to 400 ml.
146.25
NH2 SO4 left
400
= 0.367
0.367
MH2SO4 left
0.183
2
Exercise 7:
Equivalents of NaOH 50 0.2 10
Equivalents of HCl 5 1 5
Equivalent of NaOH left after reaction with HCl = 10 – 5 = 5
Also, NaOH reacts with FeCl3 = Equivalent of Fe(OH)3 = Equivalent of Fe2O3
15 0.1 1.5
Equivalent of NaOH left finally = 5 – 1.5 = 3.5
3.5
NNaOH left
0.05N
Total volume 70 litre
70
Also, equivalent of Fe2O3 = 1.5
W
1.5
M
6
1.5 160
WFe2 O3
40 g
6
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
33
Exercise 8:
Exercise 9:
We know, normality of the solution is given by
5
NAgNO3
0.3059
100
170
1.04 1000
Now, chloride samples contain different type of combination of NaCl, KCl and
NH4Cl or as individual component in each 0.3 g sample. Minimum volume of
AgNO3 which will give rise to complete precipitation in each case, will be the
maximum value of Meq. of chloride in a sample, i.e., for NH4Cl only as
independent constituent, NH4Cl has minimum equivalent weight of all the three.
meq. of AgNO3 = meq. of NH4Cl
0.3
0.3059 V
1000 V 18.33 ml
53.5
change in ON 2
HSO3
SO24 or HSO4
4
6
6
HSO3
0.017 M
= 2 0.017 N = 0.034
In the first suppose the ON of Mn in the product is X
0.01 M MnSO4 = 0.01 (7 – X)N MnO4
m.e. of HSO3 = m.e. of MnO4
0.034 25 = 0.01(7 – X)16.9
0.034 25
7X
5.00
16.9 0.01
or X = 2
Now in the second titration, suppose the ON of Mn in the product is Y.
0.01 M MnO4 = 0.01(7 – Y)N MnO4
0.034 25 = 0.01(7 – Y) 28.6
0.034 25
7Y
3
0.01 28.6
Y=4
Exercise 10:
(a) The titration reactions in this case are
Na2 CO3 HCl NaHCO3 NaCl and NaOH HCl NaCl H2 O
Thus, we have
m.e. of Na2CO3 + m.e. of NaOH = m.e. of v1 mL (say) of N HCl
1
1
1000
1000 1 v1; v1 34.4 mL
106
40
(b) The reactions in this case are,
Na2 CO3 HCl NaHCO3 NaCl
NaHCO3 HCl NaCl H2 O CO2
produced
NaHCO3 HCl NaCl H2 O CO 2
originally present
and NaOH HCl NaCl H2 O
Thus, we have,
m.e. of Na2 CO3 m.e. of NaHCO3 m.e. of NaHCO3 m.e. of NaOH
produced
originally present
= m.e. of v2 mL (say) of N HCl
1
1
1
1
1000
1000
1000
1000 1 v 2
106
106
84
50
v2 = 55.8 mL
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
34
(c) The reactions in this case are,
NaHCO3 produced HCl
NaCl H2 O CO2
and NaHCO3 originally present HCl NaCl H2O CO2
Thus we have,
m.e. of NaHCO3 m.e. of NaHCO3 m.e. of v 3 mL say of N HCl
produced
originally present
or m.e. Na 2CO3 m.e. of NaHCO3 m.e. of v 3 mL say of N HCl
1
1
1000
1000 1 v 3
106
84
v3 = 21.3 mL
Exercise 11:
Let atomic weights of P and Q be ‘a’ and ‘b’ respectively.
Molecular weight of P2Q3 = 2a + 3b and molecular weight of PQ2 = a + 2b
2a 3b 0.15 15.9 , and
a 2b 0.15 9.3
Thus, a = 26, b = 18
Exercise 12:
CaCO3 2HCl CaCl2 H2O CO 2
44 gm CO2 100 g CaCO3
0.88 g CO2
% purity
Exercise 13:
100 0.88
2.0 g CaCO3
44
2
100 50%
4
2 mole H2 1 mole O2 2 mole H2O
1 mole O2 2 mole H2O
1.57 mole O2 2 1.57 mole H2O 3.14 mole H2 O
2 mole of H2 O
or, moles of H2O 1.57 mole of O2
= 3.14 mole
1 mole of O2
Exercise 14:
Balanced equation is 2VO
Initialmoles
Final moles
3Fe2O3
6FeO
V2O5
2
5.75
0.0298
0.0359
0
0
67
160
0.0359 2
(0.0359 2) (0.0359 1)
0.0298
0
3
Moles of FeO formed 0.0359 2
Weight of FeO formed 0.0359 2 72 5.18 g
Here, the limiting reagent is Fe2O3.
Exercise 15:
112
5 103 5 10 3
22400
Let equivalent weight of metal be E, then equivalents of metal = equivalents of H2
0.275
moles of H2 2
E
2 5 10 3
E 27.52
Number of moles of H2(n)
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
35
SOLVED PROBLEMS
Subjective:
‘0’ LEVEL
Illustration 1. The reaction
2C O 2 2CO
is carried out by taking 24 g of carbon and 96 g of O 2 .
Find out:
(i) Which reactant is left in excess?
(ii) How much of it is left?
(iii) How many moles of CO are formed?
(iv) How many grams of other reactant should be taken so that nothing is left at
the end of reaction?
Solution:
2C O2 2CO
24
96
12
32
Mole after reaction
0
2
Mole ratio of C : O2 : CO :: 2 : 1: 2
(i) O 2 is left in excess.
Mole before reaction
2
(ii) 2 moles of O 2 or 64 g of O 2 is left.
(iii) 2 moles of CO of 56 g or CO is formed.
(iv) To use O 2 completely, total 6 moles of carbon or 72 g of carbon is needed.
Prob 2.
Calculate the mass of 90% pure MnO2 to produce 35.5g of Cl2 according to the
following reaction.
MnO2 + 4HCl MnCl2 + Cl2 + 2H2O
Sol.
MnO2 + 4HCl MnCl2 + Cl2 + 2H2O
87g
71g
71g Cl2 is produced by 87g of MnO2
35.5g Cl2 is produced =
87 35.5
= 43.5g
712
90g pure MnO2 is present in 100g sample
100 43.5
43.5g pure MnO2 =
= 48.3g
90
Prob 3.
8 gm of methane is burnt with 4.48L of O2 at STP. Find out the volume of CO2 gas
produced at STP and also the weight of CO2 gas.
Sol.
The balanced chemical
CH4
+ 2O2
1 mol
2 mol
16 gm
2 22.4 L
equation is
CO2
+ 2H2O
1 mol
22.4 L
44 gm
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
36
8
0.5 mol
16
4.48L
No. of moles of O2 =
0.2mol
22.4L
Now since 1 mole of CH4 requires 2 mol (i.e. 44.8 L) of O2 for complete combustion.
But the given moles of O2 is only 0.2 mol. So, O2 is the limiting reagent.
Again, since 2 moles of O2 reacts with 1 mol of CH4 to give 22.4 L of CO2
at STP.
So 0.2 mole of O2 will react with 0.1 mol of CH4 to give 2.24 L of CO2.
Wt. of CO2 produced
= 0.1 mol 44
= 4.4 gms of CO2
No. of moles of CH4 =
Prob 4.
Calculate the volume of hydrogen liberated at 27°C and 760 mm pressure by treating
1.2g of magnesium with excess of hydrochloric acid.
Sol.
The balanced equation is
Mg + 2HCl = MgCl2 + H2
1 mole
1 mole
24g
22.4 litre at NTP
24g of Mg liberate hydrogen = 22.4 litre
22.4
1.2g of Mg will liberate hydrogen =
1.2 = 1.12 litre
24
Volume of hydrogen under given condition can be calculated by applying
P1V1 P2 V2
T1
T2
P1 = 760 mm
P2 = 760 mm
T1 = 273 K
T2 = (27 + 273) = 300K
V1 = 1.12 litres
V2 = ?
760 1.12 300
V2 =
1.2308litres
273
760
Prob 5.
40 ml N/10 HCl and 60 ml N/20 KOH are mixed together. Calculate the normality of
the acid or base left. What is the normality of the salt formed in the solution?
Sol.
Milli equivalents of HCl = N × V (ml) =
1 40
4
10
1 60
Milli equivalents of KOH = N × V (ml) =
3
20
One milli equivalent of an acid neutralizes one milli equivalent of a base
Milli equivalent of HCl left = 4 – 3 = 1
Total volume of the solution = 40 + 60 = 100 ml
Milli equivalents of HCl = N × V (ml)
1 = N × 100
Normality (N) of HCl left in solution = 0.01
Salt formed = Milli equivalent of acid or base neutralized
Milli equivalents of the salt formed = N × V (ml)
3 = N × 100
Normality (N) of salt formed = 0.03
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
37
LEVEL – I
Prob 1.
A solution of H2 O2 , labelled as ‘20 volumes’, was left open. Due to this some
H2 O2 decomposed and the volume strength of the solution decreased. To determine
the new volume strength of the H2 O2 solution, 10 mL of the solution was taken and it
was diluted to 100 mL. 10mL of this diluted solution was titrated against 25 mL of
0.0245 M KMnO 4 solution under acidic conditions. Calculate the volume strength of
the H2 O2 solution.
Sol.
If a solution of H2 O2 has normality = N, it means that 1 litre of the solution contains N
equivalents.
N
1 mL of it would contain
equivalents.
1000
1
N
moles of H2 O2 in 1 mL =
2 1000
1 1
N
Moles of O 2 it gives =
2 2 1000
1 1
N
Volume of O 2 at STP in mL = 22400
5.6 N
2 2 1000
Volume strength = 5.6 Normality
0.0245 25 5
= 0.306
10
Normality of the 100mL H2 O2 solution = 0.306
(that has been diluted)
Normality of the original 10 mL H2 O2 solution = 0.306 10 = 3.06
Volume strength = 5.6 3.06 = 17.136
Normality of H2O2 solution =
Prob 2.
A solution contains Na2 CO3 and NaHCO3 . 20 cm3 of this solution require 5.0 cm3 of
0.1M H2 SO4 solution for neutralization using phenolphthalein as the indicator. Methyl
orange is then added when a further 5.0 cm3 of 0.2 M H2 SO4 was required. Calculate
the masses of Na2 CO3 and NaHCO3 in 1 L of this solution.
Sol.
1
1
Na2CO3 + H2 SO 4 NaHCO3 Na2 SO 4
2
2
phenolphthalein would change the colour after this reaction.
Meq. of H2SO4 used for 5 ml mixture using phenolphthalein as indicator
= 2 0.1 5 = 1
1
Meq. of Na2 CO3 = 1
2
Now methyl orange is added in this solution after 1 end point
Meq. of H2 SO4 used for solution after 1 end point using methyl orange as indicator =
5 0.2 2 = 2
1
Meq. of Na2 CO3 + Meq. of NaHCO3 = 2
2
Meq. of NaHCO3 = 2 -1 = 1
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
38
W
1000 1 W 0.084g
84
0.084 1000
= 4.2 g
20
W
106
Meq. of Na2 CO3 = 2
1000 = 2 or W =
= 0.106
53
1000
0.106
Weight of NaHCO3 in 1 litre
1000 = 5.3 g
20
Weight of NaHCO3 in one litre =
Prob 3.
A 50.0 cm3 portion of a mixture of H2 SO4 and H2 C2O 4 required 48.9 cm3 of 0.15 M
NaOH solution for titration. Another 50 cm3 required 38.9 cm3 of 0.10 N
KMnO 4 solution for titration. Calculate the masses of H2 SO4 and H2 C2 O4 present per
dm3 of the solution.
Sol.
H2 SO 4 and H2 C2 O 4 will react with NaOH since they are acids. Both will have ‘n’
factor 2 when they react with NaOH. Since S in H2 SO4 is already in its maximum
oxidation state of +6, H2 SO4 will not react with KMnO 4 and only provides acidic
medium for the reaction.
Let the moles of H2 SO 4 and H2 C2O 4 be a and b respectively in 50 cc
solution
48.9 0.15
2a + 2b =
=7.335 10-3
1000
38.9 0.1
2b =
= 3.89 10-3
1000
b = 1.945 10-3 a = 1.7225 10-3
mass of H2 SO 4 in 1 L = 1.7225 10-3 20 98 = 3.3761 g
mass of H2 C2O 4 in 1 L = 1.945 10-3 20 90 = 3.501 g
Prob 4.
20 g of a sample of Ba(OH)2 is dissolved in 10 ml. of 0.5N HCl solution. The excess
of HCl was titrated with 0.2N NaOH. The volume of NaOH used was 10 cc. Calculate
the percentage of Ba(OH)2 in the sample.
Sol.
Milli eq. of HCl initially = 10 0.5 = 5
Milli eq. of NaOH consumed = Milli eq. of HCl in excess
= 10 0.2 = 2
Milli eq. of HCl consumed = Milli eq. of Ba(OH)2
= 5-2=3
eq. of Ba(OH)2 = 3/1000 = 3 10-3
Mass of Ba(OH)2 = 3 10-3 (171/2) = 0.2565 g.
% Ba(OH)2
=
(0.2565/20) 100 = 1.28%
Prob 5.
A sample of 2 g containing Na2CO3 and NaHCO3 loses 0.248 g when heated to
300 oC, the temperature at which NaHCO3 decomposes to Na2CO3, CO2 and H2O.
What is the percentage of Na2CO3 in the given mixture?
Sol.
The loss in weight is due to removal of CO 2 and H2 O which escape out on heating.
Weight of Na2 CO3 in the product = 2.00 – 0.248 = 1.752 g
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
39
Let weight of Na2 CO3 in the mixture be x g
Weight of NaHCO3 2.00 x g
Since Na2 CO3 in the products contains x g of unchanged reactant Na2 CO3 and rest
produced from NaHCO3.
The weight of Na2 CO3 produced by NaHCO3 = (1.732 x)
NaHCO3
Na2 CO3 H2O CO 2
2.0 x
1.752 x
Applying POAC for Na atom,
1 moles of NaHCO3 2 moles of Na2 CO3
2 x
84
2
1.752 x
106
x 1.328 g
% of Na2 CO3
1.328
100 66.4%
2
Prob 6.
A mixture of KBr, NaBr weighing 0.56 gm was treated with aqueous solution of Ag+
and the bromide ion was recovered as 0.97 gm of pure AgBr. What was the weight of
KBr in the sample?
Sol.
KBr
+
NaBr
+
Ag+
AgBr
a gm
(0.56 – a) gm
0.97 gm
Applying POAC for Br atoms,
Moles of Br in KBr + Moles of Br in NaBr = Moles of Br in AgBr
or 1 Moles of KBr + 1 Moles of NaBr = 1 Moles of AgBr
a
(0.56 a) 0.97
(MKBr = 199, MNaBr = 103, MAgBr = 188)
119
103
188
a = 0.1332 gm
0.1332
Percentage of KBr in the sample =
100 = 23.78
0.560
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
40
LEVEL – II
Prob 1.
A mixture of H2 C2O 4 (oxalic acid) and NaHC2 O 4 was dissolved in water and the
solution made upto one litre. Ten millilitres of the solution required 3.0 mL of 0.1 N
sodium hydroxide solution for complete neutralization. In another experiment 10.0 mL
of the same solution, in hot dilute sulphuric acid medium, required 4.0 mL of 0.1N
KMnO 4 solution for complete reaction. Calculate the masses of H2 C2O 4 and
NaHC2 O 4 in the mixture.
Sol.
Let the mass of H2 C2O 4 and NaHC2 O 4 in 1L solution be a and b gms respectively.
mass of H2 C2O 4 and NaHC2 O 4 in 10 mL solution will be a/100 and b/100 gms
respectively.
3 0.1
Equivalents of NaOH solution =
3 104 .
1000
Since NaOH is a base, H2 C2O 4 and NaHC2 O 4 react with it like acids and their ‘n’
factors respectively are 2 and 1.
equivalents of H2 C2O 4 + eq. of NaHC2 O 4 = 3 10 -4
a2
b
3 104
--(1)
90 100 100 112
(because molecular weight of H2 C2O 4 is 90 and NaHC2 O 4 is 112)
4 0.1
4 10 4
1000
Since KMnO 4 is an oxidising agent, it oxidises C+3 to C+4 (CO2 ) .Therefore the ‘n’
Equivalents of KMnO 4 solution =
factor of both H2 C2O 4 and NaHC2 O4 are 2.
a
b
2
2
4 10 4
90 100
100 112
b
(2) — (1) gives
10 4
112 100
b = 1.12 gms
a = 0.9 gms.
Prob 2.
------
(2)
A solution of 0.2 g of a compound containing Cu+2 and C2 O42 ions on titration with
0.02 M KMnO 4 in presence of H2 SO4 consumes 22.6 mL of the oxidant. The resultant
solution is neutralized with Na2 CO3 , acidified with dilute acetic acid and treated with
excess KI. The liberated iodine requires 11.3 mL of 0.05 N Na2 S2O3 solution for
complete reduction. Find out the molar ratio of Cu+2 to C2 O42 in the compound. Write
down the balanced redox reactions involved in the above titration.
Sol.
The mixture of Cu2 and C2 O24 are reacting separately first with KMnO 4 solution and
then with solid KI to liberate iodine. It can be seen that Cu+2 cannot be oxidized
and C2 O42 cannot be reduced. This is because Cu is already in its highest oxidation
state +2 and C cannot have an oxidation state less than +3 when it is combined with
oxygen.
0.02 5 22.6
Equivalents of KMnO 4 solution =
2.26 10 2
1000
2.26 10 3
mole of C2 O42 =
1.13 10 3
2
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
41
This is because only C2 O42 is oxidised by KMnO 4 to CO2 (‘n’ factor 2)
Equivalents of Na2 S2O3 solution =
0.05 11.3
5.65 104
1000
5.65 104
5.65 104
1
This is because only Cu+2 is reduced by KI to Cu+
5.65 104
mole ratio of Cu+2 to C2 O42 =
= 0.5
1.13 10 3
Reactions: 2Mn O 4 + 5C2 O42 + 16H+ 2Mn+2 +10CO2 + 8H2 O
moles of Cu2
(2Cu+2 + 4I- Cu2I2 +I2 ) , I2 +2S2 O3-2 2I- + S4 O62-
Prob 3.
A 2.0 g sample of a mixture containing sodium carbonate, sodium bicarbonate and
sodium sulphate is heated till the evolution of CO 2 ceases. The volume of CO 2 at 750
mmHg pressure and at 298 K is measured to be 123.9 mL. A 1.5 g of the sample
requires 150 mL of M/10 HCl for complete neutralization. Calculate the percentage
composition of the components of the mixture.
Sol.
Out of Na2 CO3 and Na2 SO4 only NaHCO3 decomposes on heating to give CO 2 gas,
according to the equation 2NaHCO3 Na2 CO3 + CO 2 + H2O
PV
750 123.9
5 103
RT 760 1000 0.082 298
moles of NaHCO3 = 2 5 10-3 = 0.01
150 101
Equivalents of HCl used =
1.5 10 2
1000
1.5
Equivalents of NaHCO3 in 1.5 g = 0.01
7.5 10 3
2
Equivalents of Na2 CO3 = 1.5 10-2 - 7.5 10-3 = 7.5 10-3
Moles of CO 2
7.5 103
(when Na2 CO3 reacts with
2
NaCl, CO 2 and H2 O . No atom undergoes change in oxidation state.
Moles
of
Na2 CO3
HCl
it
gives
‘n’ factor of Na2 CO3 = 2) = 3.75 10-3
Mass of NaHCO3 in 1.5 g = 7.5 10-3 84 = 0.63 g
Mass of Na2 CO3 in 1.5 g = 3.75 10-3 106 = 0.3975 g
mass of Na2 SO 4 = 1.5 - 0.63 - 0.3975 = 0.4725 g.
Percentage of NaHCO3 = 0.63 100 = 42%
1.5
0.3975
Percentage of Na2 CO3 =
100 = 26.5%
1.5
0.4725
Percentage of Na2 SO4 =
100 = 31.5%
1.5
Prob 4.
A 1.0 g sample of Fe 2O3 solid of 55.2% purity is dissolved in acid and reduced by
heating the solution with zinc dust. The resultant solution is cooled and made upto
100.0 mL. An aliquot of 25.0 mL of this solution requires 17.0 mL of 0.0167 M
solution of an oxidant for titration. Calculate the number of electrons taken up by the
oxidant in the reaction of the above titration.
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
42
Sol.
Prob 5.
55.2
1 = 0.552 g
100
0.552
Moles of Fe2O3 =
3.45 103
160
n 0.0167 17
Equivalents of the oxidant =
2.84 10 4 n
1000
(n is the ‘n’ factor of the oxidant)
Since on adding Zinc dust to the Fe 2O3 solution all the Fe+3 will become Fe+2 , moles
Mass of Fe 2O3 =
of Fe2+ in 100 mL =3.45 10-3 2 = 6.9 10-3
Equivalents of Fe2+ in the 25 mL that is reacting with oxidant
6.9 103
=
1.725 10 3
4
according to the Law of Equivalents =1.725 10-3 = 2.84 10-4 n
1.725 103
n=
= 6.07 6
2.84 10 4
A 8.0 g sample contained Fe3 O 4 , Fe 2O3 and inert materials. It was treated with an
excess of aqueous KI solution in acidic medium, which reduced all the ion to
Fe+2 ions. The resulting solution was diluted to 50.0 cm3 and a 10.0 cm3 of it was
taken. The liberated iodine in this solution required 7.2 cm3 of 1.0 M Na2 S2O3 for
reduction to iodide. The iodine from another 25.0 cm3 sample was extracted, after
which the Fe+2 ions were titrated against 1.0 M MnO 4 in acidic medium. The volume
of K MnO 4 solution used was found to be 4.2 cm3 . Calculate the mass percentages
of Fe3 O 4 and of Fe2 O3 in the original mixture.
Sol.
This problem can be done by two methods.
In the first method, we break up Fe3 O 4 as an equimolar mixture of FeO and Fe 2O3 .
Method (1)
Fe3 O 4 is FeO. Fe2 O3
1 7.2
7.2 103
1000
Equivalents of I2 in 10 cc
= 7.2 10-3
Equivalents of Na2 S2O3 =
Equivalents of I2 in 50 cc
= 7.2 10-3 5 3.6 102
Since equivalents of I2 is equal to that of KI which in turn is equal to the total
equivalents of Fe 2O3 ( Fe 2O3 in the free state and Fe 2O3 combined with FeO).
equivalents of total Fe 2O3
= 3.6 10-2
4.2 1 5
2.1 102
1000
Since KMnO 4 reacts with the total Fe2+ ( Fe2+ in FeO and Fe2+ that was produced by
the action of KI on Fe 2O3 )
Equivalents of KMnO 4 solution =
equivalents of total Fe+2 in 50 mL = 2.1 10 -2 2 = 4.2 10-2
Since equivalents of Fe+2 produced from Fe 2O3 is equal to the equivalents of Fe 2O3
Equivalents of FeO = 4.2 10-2 - 3.6 10 -2 6 103
moles of FeO = 6 10-3
moles of Fe 2O3 combined with FeO = 6 10 -3
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
43
3.6 10 2
2
(because when Fe2O3 Fe2+ ‘n’ factor is 2). = 1.8 10-2
total moles of Fe 2O3 =
moles of Fe 2O3 in the free state = 1.8 10-2 - 6 10-3 = 1.2 10-2
mass of Fe3 O 4 = 6 10-3 232 = 1.392 g
mass of Fe 2O3 =1.2 10-2 160=1.92 g
percentage Fe2O 4 = 17.4%
percentage of Fe 2O3 = 23.75%
Method 2:
Here we take Fe3 O 4 as a single entity.
7.2 1
7.2 103
1000
Equivalents of I2 in 50 cc = 7.2 10-3 5 = 3.6 10-2
Equivalents of Fe3 O 4 + Fe2 O3 = 3.6 10-2
Let us assume that the moles of Fe3 O 4 is x g and that of Fe 2O3 is y g. Since on
Equivalents of Na2 S2O3 =
reacting with KI both Fe3 O 4 and Fe 2O3 give Fe2+ ‘n’ factor for both is two.
2x + 2y = 3.6 10-2 -------- (1)
4.2 1 5
Equivalents of KMnO 4 =
2.1 102
1000
Equivalents of Fe2+ in 50 mL = 2.1 10-2 2 = 4.2 10-2
moles of Fe2+ in 50 mL = 4.2 10-2
Since the moles of Fe3 O 4 are x, moles of Fe2+ produced from Fe3 O 4 will be 3x and
that produced from Fe2O3 will be 2y.
3x + 2y = 4.2 10-2 --------- (2)
(2) - (1) gives x = 6 10 -3
y = 1.2 10-2
Solving this percentage of Fe3 O 4 is 17.4% and Fe2 O3 is 23.75%.
Prob 6.
A sample of hard water contains 96 ppm of SO 42 and 183 ppm of HCO3, with
Ca+2 as the only cation. How many moles of CaO will be required to remove HC O 3
from 1000 kg of this water? If 1000 kg of this water is treated with the amount of CaO
calculated above, what will be the concentration (in ppm) of residual Ca+2 ions?
(Assume CaCO3 to be completely insoluble in water). If the Ca+2 ions in one litre of
the treated water are completely exchanged with hydrogen ions, what will be its pH?
(one ppm means one part of the substance in one million part of water, mass/mass.)
+
(Given pH = log[H ])
Sol.
In 1000 kg of water the mass of HC O 3 = 183 g
183
96
= 3; moles of SO 24 =
=1
61
96
Total moles of Ca 2+ in the solution = 1 + 1.5 = 2.5
Ca(HCO3 )2 + CaO 2CaCO3 + H2 O
moles of HC O 3 =
Moles of CaO required to be added to remove all HC O 3 = 1.5. Now the Ca 2+ in
the solution will be only associated with SO 24 . Therefore moles of Ca 2+ left in the
solution = 1.
ppm of Ca 2+ = 1 40 = 40 ppm ; moles of Ca 2+ ions in 1 L of H2 O = 10-3
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
44
moles of H+ ions that Ca+2 will exchange with = 2 10-3
pH = -log(2 10-3 ) = 2.7
Prob 7.
One litre of a mixture of O 2 and O3 at STP was allowed to react with an excess of
acidified solution of KI. The iodine liberated required 40 mL of M/10 sodium
thiosulphate solution for titration. What is the mass percent of ozone in the mixture?
Ultraviolet radiation of wavelength 300 nm can decompose ozone. Assuming that
one photon can decompose one ozone molecule, how many photons would have
been required for the complete decomposition of ozone in the original mixture.
Sol.
Total moles of the mixture =
1
0.0446 4.46 10 2
22.4
40
1
Equivalents of Na2 S2O3 solution =
4 10 3
1000 10
equivalents of I2
= 4 10-3
equivalents of KI reacted = 4 10-3
equivalents of O 3
= 4 10-3
When O 3 reacts with KI it converts to O 2 and O-2
the ‘n’ factor for O 3 in this reaction is 2
4 10 3
2 103
2
moles of O 2 = 4.46 10-2 - 2 10-3 = 4.26 10-2
mass percent of ozone in the mixture
2 103 48
=
100 = 6.57%
2 103 48 4.26 102 32
number of O 3 molecules = 2 10-3 6.023 1023 = 1.2 1021
moles of O 3 =
number of photons required = 1.2 10 21
Prob 8.
0.2 gm of a mixture of NaOH and Na2CO3 and inert impurities was first titrated with
phenolphthalein and N/10 HCl, 17.5 ml of HCl was required at the end point. After
this methyl organge was added and 2.5 ml of same HCl was again required for next
end point. Find out percentage of NaOH and Na2CO3 in the mixture.
Sol.
Let, W 1 gm NaOH and W 2 gm Na2CO3 was present in mixture. At phenophthalein end
point,
1
W1 1 W2
17.5 10–3
… (1)
=
40 2 53 10
At second end point following reaction takes place
1
Eq. of NaHCO3 = Eq. of HCl used (in second titration) =
Eq. of Na2CO3
2
1 W2
1
= 25
10–3
2 53
10
W 2 = 0.0265 gm
Putting the value of W 2 in Eq. (1), we get
W 1 = 0.06 gm
0.06
Percentage of NaOH =
100 = 30%
0.2
0.0265
Percentage of Na2CO3 =
100 = 13.25%
0.2
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
45
Objective:
‘0’ LEVEL
True and False:
Prob 1.
The oxidation number of Ni in Ni(CO)4 is +2.
Sol.
False
Prob 2.
In the reaction, F + 1 O OF , oxygen is an oxidant.
2
2
2
2
Sol.
False
Prob 3.
The oxidation state of Cr in CrO5 is +10.
Sol.
False
Prob 4.
Cu+ undergoes disproportionation to Cu and Cu++ ions.
Sol.
True
Prob 5.
2H 2 O 2 2H 2 O + O 2 is an example of disproportionation.
Sol.
True
Fill in the Blanks:
Prob 6.
The element which shows highest oxidation number is ……………………..
Sol.
(Os in OsO4)
Prob 7.
A redox reaction simply involves the transfer of electrons from ……………. to
………………………………
Sol.
(reductant, oxidant)
Prob 8.
The lowest possible oxidation state of nitrogen is ……………………
Sol.
-3, N
Prob 9.
Maximum number of moles of PbSO4 that can be precipitated by mixing 20 ml of
0.1 M Pb(NO3)2 and 30 ml of 0.1 M Na2SO4 is …………………
Sol.
(0.002)
3-
Prob 10. Of the halide ion …………… is the most powerful reducing agent.
Sol.
(I–)
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
46
LEVEL – I
MCQ Single Correct:
Prob 1.
Find the weight of KOH in its 50 milli equivalents
(A) 1.6
(B) 2.2
(C) 2.8
(D) 4.8
Sol.
Meq =
Prob 2.
The number of neutrons in a drop of water (20 drops = 1mL) at 4°C
(A) 6.023 1022
(B) 1.338 1022
(C) 6.023 1020
(D) 7.338 10 22
Sol.
Mass of a drop of water = 0.05 1 g = 0.05 g
0.05
No. of moles of water =
18
0.05
No. of water molecules =
6.023 1023
18
1 water molecule certain 8 neutrons
0.05
0.05 8
6.023 1023 molecule certain
6.023 1023 neutrons
18
18
= 0.1338 1023 = 1.338 1022
(C)
Prob 3.
Weight of 1 atom of an element is 6.644 10-23 g. What is number of atoms of element
in 40 kg of it?
(A) 103 g atom
(B) 102 g atom
(C) 10 4 g atom
(D) 10 g atom
Sol.
Weight of Avogadro number (N) of atoms of the element
= 6.644 10-23 6.023 10 23 = 40 g
40 g = weight of 1g atom
40 103 g = weight of 103 g atom
(A)
Prob 4.
A compound contains 3.2% of oxygen. The minimum mol wt. of the compound is
(A) 300
(B) 440
(C) 350
(D) 500
Sol.
The compound must contain at least one oxygen atom.
So, a minimum of 1 g atom of oxygen will be present in 1 g molecule i.e., 1 mole of the
compound.
If M is the mol.wt. of the compound then since 16 is the atomic mass of oxygen so
minimum of 16 g of oxygen will be present in M g of the compound
16
Thus, % of oxygen =
100
M
16 100
or 3.2 =
or M = 500
M
(D)
weight
weight
1000 50 =
1000
Eq.wt.
56
Weight of KOH = 2.80 g
(C)
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
47
Prob 5.
No. of oxalic acid molecules in 100 ml of 0.02 N oxalic acid are
(A) 6.023 1020
(B) 6.023 1021
22
(C) 6.023 10
(D) 6.023 1023
Sol.
Normality = Molarity Valence factor
Normality
Molarity =
Valence factor
mol. wt. of oxalic acid
Valence factor for oxalic acid =
Eq. wt. of oxalic acid
02
Molarity =
=0.01
2
Number of millimoles = 0.01 100
Number of moles = 0.001
No. of oxalic acid molecules = 0.001 6.023 1023 =6.023 1020
(A)
Prob 6.
112 ml of a gas is produced at S.T.P. by the action of 412 mg of alcohol ROH
with CH3MgI . The molecular mass of alcohol is
(A) 32 g
(B) 41.2 g
(C) 82.4 g
(D) 156 g
Sol.
ROH+ CH3MgI CH4 + Mg
1 mole
1 mole
So the gas produced is CH4 .
1 mole CH4 will be produced from 1 mole of alcohol
22.4 lit CH4 will be produced by mol.wt. of alcohol
112 ml CH4 is produced from 4.12 mg of alcohol
412 22400
22400 ml CH4 is produced from 0
mg = 82400 mg = 82.4 g
112
So Mol.wt. of alcohol = 82.4 g
(C)
Prob 7.
If 0.5 mole of BaCl2 is mixed with 0.2 mol of Na3PO 4 , the maximum amount of
Ba3 (PO 4 )2 that can be formed is:
(A) 0.7 mol
(B) 0.5 mol
(C) 0.2 mol
(D) 0.1 mol
Sol.
Let us first solve this problem by writing the complete balanced reaction.
3BaCl2 + 2 Na3PO4 Ba3 (PO 4 )2 + 6NaCl
3
times the moles of Na3PO 4 .
2
So, to react with 0.2 mol of Na3PO 4 , the moles of BaCl2 required would be
We can see that the moles of BaCl2 used is
3
= 0.3. Since BaCl2 is 0.5mol, we can conclude that Na3PO 4 is the limiting
2
1
reagent. Therefore, moles of Ba3 (PO 4 )2 formed is 0.2 = 0.1 mol.
2
(D)
Alternatively,
0.2
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
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We can use the concept of equivalents to arrive at the answer.
BaCl2 and Na3PO 4 must be reacting without undergoing any change in oxidation state
since in Ba3 (PO 4 )2 the oxidation states of Ba, P and O remain fixed at +2, +5 and –2
respectively and the other product is expected to be NaCl.
Equivalents of BaCl2 = 0.5 2 (Calculate 'n' factor as number of moles of cation in 1
mole of substance oxidation state of each cation).
Equivalents of Na3PO 4 = 0.2 3 = 0.6
Therefore, Na3PO4 is the limiting reagent.
Equivalents of Ba3 (PO 4 )2 formed = 0.6
'n' factor of Ba3 (PO 4 )2 = 6 ( 'n' factor of Na3PO 4
Na3PO4 and Ba3 (PO 4 )2 will be in the molar ratio of 2:1)
Moles of Ba3 (PO 4 )2 =
0.6
= 0.1
6
(D)
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LEVEL – II
MCQ Single Correct:
Prob 1.
A 10.0 g sample of a mixture of calcium chloride and sodium chloride is treated with
Na2 CO3 to precipitate the calcium as calcium carbonate. This CaCO3 is heated to
convert all the calcium to CaO and the final mass of CaO is 1.62 gms. The % by mass
of CaCl2 in the original mixture is
(A) 15.2%
(B) 32.1%
(C) 21.8%
(D) 11.07%
Sol.
Moles of CaO =
Prob 2.
Equal volumes of 1 M each of KMnO 4 and K 2 Cr2 O7 are used to oxidise Fe(II) solution
in acidic medium. The amount of Fe oxidized will be
(A) more with KMnO 4
(B) equal with both oxidizing agents
(C) more with K 2 Cr2 O7
(D) cannot be determined
Sol.
The 'n' factor of KMnO 4 is 5 while that of K 2 Cr2 O7 is 6. So for the same number of
moles, K 2 Cr2 O7 will have greater equivalence than KMnO4 .
(C)
Prob 3.
How many millilitre of 0.5 M H2 SO4 are needed to dissolve 0.5 g of Cu(II) carbonate?
(A) 6.01
(B)4.5
(C) 8.1
(D) 11.1
Sol.
1.62
1.62
Moles of CaCl2
56
56
1.62
3.21
Mass of CaCl2
111 = 3.21 gm % =
100 = 32.1%
56
10
(B)
0.5 2 1000
M
(Eq. wt. of CuCO3 = )
123.5
2
V =8.097 = 8.1
(C)
0.5 2 V =
Prob 4.
NH 2
NO 2
Equivalent wt. of Aniline is
(A) M/4
(C) M/6
Sol.
(B) M/5
(D) M/ 8
(C)
Prob 5.
The mass of BaCO3 produced when excess CO2 is bubbled through a solution of 0.205
mol Ba(OH)2 is
(A) 81 g
(B) 40.5 g
(C) 20.25 g
(D) 162 g
Sol.
Ba(OH)2 + CO2 BaCO3 + H2O
Atomic wt. of BaCO3 = 137 + 12 + 16 3 = 197
No. of mole =
wt. of substance
mol wt.
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1 mole of Ba(OH)2 gives 1 mole of BaCO3
205 mole of Ba(OH)2 will give .205 mole of BaCO3
wt. of 0.205 mole of BaCO3 will be
0.205 197 = 40.385 gm 40.5 gm
(B)
Prob 6.
The number of water molecules present in a drop of water (volume 0.0018 ml) at room
temperature is
19
18
(A) 6.023 10
(B) 1.084 10
17
(C) 4.84 10
(D) 5.023 1023
Sol.
Density =
Mass
g
;1
or g = ml
Volume
ml
0.0018 ml = 0.0018 gm
No. of moles =
weight
0.0018
1 104
Molecular weight
18
No. of water molecules = 6.023 1023 1 10-4 = 6.023 1019
(A)
Prob 7.
What will be the volume of CO2 at NTP obtained on heating 10 grams of (90% pure)
limestone?
(A) 22.4 litre
(B) 2.016 litre
(C) 2.24 litre
(D) 20.16 litre
Sol.
CaCO3 CaO +
10 gm
90% pure 9 gm =
CO2
9
mole
100
CaCO3 CO2 = 0.09 mole
At NTP vol. CO2 = 0.09 22.4 = 2.016 L
(B)
Prob 8.
55 g Ba(MnO4)2 sample containing inert impurity is completely reacting with 100 mL of
56 ‘vol’ strength of H2O2, then what will be the percentage purity of Ba(MnO4)2 in the
sample ? (Ba – 137, Mn – 55)
(A) 40%
(B) 25%
(C) 10%
(D) 68.18%
Sol.
m eq. of Ba(MnO4)2 reacted = m eq. of H2O2 reacted =
56 100
5.6
= 1000 m eq. of H2O2 1 eq. of H2O2
moles of Ba(MnO4)2 (n-factor = 10) = 0.1 mole
wt. of Ba(MnO4)2 = 0.1 375 g = 37.5 g
% purity of Ba(MnO4)2 =
37.5
100 68.18%
55
(D)
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MCQ Multiple Correct Type:
Prob 9.
Which of the following have the numerical value ‘6’ as the answer?
(A) oxidation state of S in H2SO5
(B) oxidation state of Cr in CrO5
(C) period to which Pb belongs
3+
(D) n-factor for K2Cr2O7 in acid medium when it forms Cr
Sol.
A, B, C, D
Prob 10. The number of molecules in 11 g of CO2 is same as that in
(A) 8 g of oxygen
(B) 16 g of oxygen
(C) 7 g of CO
(D) 3.5 g of oxygen
Sol.
A, C
Prob 11. 25 ml of 0.50 M H2O2 solution is added to 50 ml of 0.20 M KMnO4 in acidic solution.
Which of the following statement(s) is/are true?
(A) 0.10 mole of oxygen is liberated
(B) 0.005 mole of KMnO4 does not react with H2O2
(C) 0.0125 gmole of oxygen gas is evolved
(D) 0.0025 mole of H2O2 does not react with KMnO4
Sol.
B, C
Prob 12. A solution of Na2S2O3 is iodometrically titrated against 0.2505 gm of KBrO3. This
process requires 90 ml of Na2S2O3 solution, the strength of Na2S2O3 isf
(A) 0.2 M
(B) 0.1 M
(C) 0.05 M
(D) 0.1 N
Sol.
B, D
Assertion – Reason Type:
The following questions consist of two statements each printed as Assertion and Reason. While
answering these questions, you are required to choose any one of the following four responses.
(A) If both Assertion and Reason are true and the Reason is a correct explanation of the
Assertion.
(B) If both Assertion and Reason are true but Reason is not a correct explanation of the
Assertion.
(C) If Assertion is true but the Reason is false.
(D) If the Assertion is false but the Reason is true.
Prob 13. (Assertion): Among KMnO4, KClO4, HNO3 only HNO3 acts as oxidant while rest two
may also acts as reductant.
(Reason): In these molecules central atom is in its highest oxidation state
Sol.
D
Prob 14. (Assertion): In iodimetry, I2 is reduced while in iodometry I are oxidized.
(Reason): Iodimetry and iodometry are important analytical methods.
Sol.
B
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Comprehension Based Questions:
Comprehension-I
Calcium carbonate is found in many everyday products, such as marble, chalk, and antacids.
Commercially, it can be excavated as either calcium oxide or calcium carbonate. Calcium oxide
converts to calcium carbonate upon exposure to carbon dioxide under high pressure. At room
temperature, calcium carbonate is relatively insoluble in water. The solubility of calcium carbonate
increases as the pH of the aqueous solution decreases, because calcium carbonate is basic.
Reaction I can be combined with reaction II to convert calcium oxide into calcium carbonate in
water.
CaO(s) + H2O(l) Ca(OH)2(aq)
Reaction – I
Ca(OH)2(aq) + CO2(g) CaCO3(s) + H2O(l)
Reaction – II
Ca(OH)2 and CaCO3 both readily form a relatively insoluble white solid precipitate in water.
Because of this low solubility, the products of both reaction I and reaction II are easy to isolate
from solution. Fortunately, calcium hydroxide is more soluble in water than is calcium carbonate,
which allows for the selective precipitation of calcium carbonate in reaction II. Because of this
industrial processes for isolating calcium carbonate are primarily water – based. Calcium
carbonate can also be formed according to the following equilibrium reaction:
CaO(s) + CO2(g)
CaCO3(s)
Reaction – III
Prob 15. Which one has least mass percent of calcium?
(A) Calcium carbonate
(B) Calcium oxide
(C) Calcium chloride
(D) Calcium fluoride
Sol.
C
Prob 15. If 10.00 grams of Ca(OH)2(s) produces 5.00 grams of CaCO3(s), what is the percent
yield for the reaction?
(A) 37%
(B) 50%
(C) 75%
(D) 100%
Sol.
A
Comprehension-II
Pyrolusite, MnO2, is the main ore from which manganese is produced. The manganese content of
the ore may be determined by reducing the MnO2 under acetic conditions to Mn2+ with the oxalate
ion, C2 O24 , the oxalate ion being oxidized to carbon dioxide during the reaction. The analytical
determination is carried out by adding a known excess volume of oxalate solution to a suspension
of the pyrolusite and digesting the mixture on a hot water bath until all the MnO2 has been
reduced. The excess, unreacted oxalate solution is then titrated with standardized potassium
permanganate, KMnO4 solution after which the manganese content of the ore can be calculated.
A student prepared a standard solution of sodium oxalate by weighing 3.2 g of the dry anhydrous
salt in distilled water and making the solution up to 500 mL. 25 mL of the oxalate solution required
24.76 mL of KMnO4 solution. (At. wt. K = 39, Mn = 55, O = 16, C = 12, Na = 23)
Prob 17. What is the molarity of KMnO4 solution?
(A) 0.04776
(C) 0.038
Sol.
(B) 0.01929
(D) 0.028
B
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Prob 18. How many moles of C2 O 42 ions will be oxidized by 1 mole MnO 4 ?
(A) 1/2
(B) 3/2
(C) 5/2
(D) 7/2
Sol.
C
Matrix Match type Question:
Prob 19.
Match the following.
List – I
(Number of Moles)
(P) 0.1 moles
(Q) 0.2 moles
(R) 0.25 moles
(S) 0.5 moles
P
Q
R
(A) 3
1
2
(B) 4
2
4
(C) 2
1
3
(D) 4
4
1
Sol.
C
Prob 20.
Match the following.
Column – I
(A) NH3 NO3
Sol.
(1)
(2)
(3)
(4)
List – II
(Amount)
4480 ml of CO2 at STP
0.1 gm atom of Fe
1.5 1023 molecules of oxygen gas
9 ml of water
S
4
2
4
2
(p)
Column – II
M.W./20
(B) Fe 2S3 O2 FeSO4 SO2
(C) CaCO3 2HCl CaCl2 H2O CO2
(q)
M.W./2
(r)
M.W./8
(D) CuS CuSO4
(s)
(t)
50
Non redox process
(A) (r); (B) (p); (C) (q, s, t); (D) (r)
Table-1
Answer questions appropriately matching the information given in the three columns of
the following table.
Column 1
Column 2
Column 3
(Amount of substance)
(Volume of gas at STP)
(Total No. of atom)
(I) 0.5 mole of SO2(g)
(i) 22.4 L
(P) 2 NA
(II) 2g of H2(g)
(ii) 33.6 L
(Q) 0.50 NA
(III) 1.5 mole of O3(g)
(iii) 11.2 L
(R) 1.5 NA
(IV) 8g of O2(g)
(iv) 5.6 L
(S) 4.5 NA
Prob 21.
Which of the following is correct combination?
(A) (I) (i) (P)
(B) (II) (iii) (P)
(C) (III) (ii) (S)
(D) (IV) (iv) (P)
Sol.
C
Prob 22.
Which of the following is incorrect combination?
(A) (I) (iii) (R)
(B) (II)(i) (P)
(C) (III) (i) (S)
(D) (IV) (iv) (Q)
Sol.
C
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Prob 23. Which of the following is correct?
(A) Volume of 8g O2(s) is equal to 2g of H2(g) at STP
(B) Volume of 1.5 mole of O3 is equal to 0.5 ml of SO2(g) at STP
(C) Volume of 1.5 mole of O3 is equal to 2g of H2(g) at STP
(D) None of these
Sol.
D
Numerical Based Question/Decimal Based Questions:
Prob 24. 1 mole of H3PO2 reacts completely to two moles of I2. I2 converts to I, what will be the
oxidation state of P in the product?
Sol.
5
Prob 25. 0.006 gms of MgSO4 was present in a 1 litre water sample (d = 1 gm/cm3). Calculate
its degree of hardness in ppm.
Sol.
5
Prob 26. An ore contains 1.34% of the mineral argentite, Ag2S, by weight. How many grams of
this ore would have to processed in order to obtain 1.00 g of pure Ag?
Sol.
85.7
Prob 27. Hydrogen evolved at NTP (in litres) on complete reaction of 27 gm of Al with excess of
aqueous NaOH would be.
Sol.
33.6
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CHAPTER PRACTICE PROBLEM (CPP)
Subjective:
LEVEL – I
1. A definite amount of impure sample of P4O6 is treated with 20ml of X (M) KMnO4 in acidic
medium to produce H3PO4 and MnCl2. 20 ml of same KMnO4 on treatment with 0.2 M FeSO4
requires exactly 10 ml of FeSO4 solution. What is the amount of pure P4O6. If 0.1 g sample is
taken calculate % purity of P4O6.
2. 20 ml solution of HIO3 is reacted with excess of an aqueous solution of SO2. The excess of
SO2 and I2 formed are removed by heating the solution. The remaining solution in neutralized
by 35.5 ml of 0.16 (N) NaOH solution. Calculate the strength of HIO3 in gm/lt.
(Round off the answer to the nearest whole number)
3. A 10 mL of K2Cr2O7 solution liberated iodine from KI solution. The liberated iodine was
titrated by 16 mL of M/25 sodium thiosulphate solution. Calculate the concentration of
K2Cr2O7 solution in milli grams per litre.
LEVEL – II
4. 1 gm of impure Na2CO3 is dissolved in water and the solution is made upto 250 ml. To
50 ml of this made up solution, 50 ml of 0.1 N HCl is added and the mixture after shaking
well, required 10 ml of 0.16 N NaOH solution for complete neutralization. Calculate % purity
of the sample of Na2CO3. (Express your answer to the nearest whole number).
5. 1.5 g of a sample of ammonium sulphate was boiled with excess of NaOH solution. Evolved
NH3 was passed in 100 mL normal solution of H2SO4. The partially neutralized acid required
160 mL N/2 NaOH for complete neutralization. Calculate % purity of ammonium sulphate
sample.
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
56
Objective:
Single correct Types:
1. 3 mol of a mixture of FeSO4 and Fe2(SO4)3 required 100 ml of 2 M KMnO4 solution in acidic
medium. Hence the mole fraction of FeSO4 in the mixture is
(A) 1/3
(B) 2/3
(C) 2/5
(D) 3/5
2. Number of moles KMnO4 that is needed to react with one mole of FeC2O4 in acidic medium is
(A) 2/5
(B) 3/5
(C) 4/5
(D) 1
3. For a given mixture of NaHCO3 and Na2CO3, volume of a given HCl required is x ml with
phenolphthalein indicator and further y ml required with methyl orange indicator. Hence
volume of HCl for complete reaction of original NaHCO3 is
x
(A) 2x
(B)
2
(C) y
(D) (y x)
4. The equivalent weight of an element is 15. It forms an acidic oxide; with potassium hydroxide
it forms a salt isomorphous with K2SO4. The atomic weight of the element would be
(A) 90
(B) 13.16
(C) 26.52
(D) 52.64
(Isomorphous salts have same molecular formula and same crystal structure)
5. A mixed solution of potassium hydroxide and sodium carbonate required 15 mL of a N/20 HCl
solution when titrated with phenolphthalein as an indicator. But the same amount of the
solution when titrated with methyl orange as an indicator required 25 mL of the same acid.
Calculate the amount of KOH present in the solution.
(A) 0.014 g
(B) 0.14 g
(C) 0.028 g
(D) 1.4 g
Multiple correct Types:
6. Which represents disproportionate?
1
2
(A) 2Cu Cu Cu
2
2
(C) Cu Zn Zn Cu
5
(B) 3I2 5I I
(D) NH4 2 Cr2 O7 N2 Cr2O3 4H2O
7. Which of the following are correct about the reaction?
FeS2 O2 Fe 2O3 SO2
(A) Equivalent weight of FeS2 is M/11
(B) Equivalent weight of SO2 is M/5
(C) S has 2 oxidaiton state in FeS2
(D) 1 mole of FeS requires 7/4 mole of O2
8. In the following reaction:
4P 3KOH 3H2O 3KH2PO2 PH3
(A) P is oxidized
(C) KOH is oxidised
(B) P is reduced
(D) P is neither oxidized nor reduced
Numerical based Type:
9. Oxidaiton state of Ni in K3Ni(CN)4 is.
10. The difference in the oxidation number of the two types of sulphur atoms in Na2S4O6 is.
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
57
ASSIGNMENT PROBLEMS
Subjective:
LEVEL - I
1.
6 10 3 g of oxygen is dissolved per kg of sea water. Calculate the parts per million (ppm) of
oxygen in sea water
2. What is effect of temperature on?
(a) Molarity (b) Molality
3. If the density of solvent is taken to be one, which one of the 1 molar and 1 molal solution is
more concentrated?
4. Differentiate between molarity and molality of a solution. When and why is molality preferred
over molarity in handling solutions in chemistry?
5. 2.82 g of glucose (mol mass=180) are dissolved in 30 g of H2O . Calculate the (i) Molality of
the solution (ii) mole fraction of glucose and water
6. A bottle of commercial sulphuric acid (density=1.787 g/ml) is labelled as 86 percent by
weight.
(i) What is the molarity of the acid?
(ii) What volume of the acid has to be used to make 1 litre of 0.2 M H2 SO4 ?
(iii) What is the molality of the acid?
7. H2O2 acts as a reductant as well as oxidant. Explain.
8. The solubility of Ba(OH)2 . 8H2 O in water at 288 K is 5.6 g per 100 g of water. What is the
molality of the hydroxide ions in the saturated solution of barium hydroxide at 288 K? (Atomic
mass: Ba = 137, O = 16, H =1)
9. What is the molarity and molality of a 13% solution (by weight) of sulphuric acid? Its density is
1.020 gm cm 3 (Atomic masses: H =1, O = 16, S = 32 amu)
10. Balance the following equations by ion electron (half reaction) method
(i) H2 S MnO4 H S Mn2 H2O
(ii) OH Br2 Br BrO3 H2 O
11. How many gm of HNO3 can be obtained from 50.0 g of KNO3 of 85% purity? (by mixing conc.
H2SO4)
12. KClO3 on heating decomposes to KCl and O2. Calculate the volume of O2 at STP liberated by
0.25 mol of KClO3.
13. Calculate amount of CO needed per g of Ni in the Mond’s process.
Ni 4CO Ni CO 4 Given at. wt. of Ni 58.7
14. What weight of CuSO4.5H2O must be taken to make 1.5 litre of 0.01 M copper (II) ion
solution?
15. Calculate the strength of K2Cr2O7 solution, whose 10 ml required 15 ml of 0.1 N Na2S2O3
solution for neutralization?
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
58
16. Calculate the weight of MnO2 and the volume of HCl of specific gravity 1.2 gml–1 and 4%
nature by weight needed to produce 1.78 litre of Cl2 at STP by the reaction.
17. A solution containing 2.68 10-3 mol of An+ ions requires 1.61 103 mol of MnO4 for the
complete oxidation of An+ to AO3 in acidic medium. What is the value of n?
18. A solution of H2O2, labeled as ’20 volumes’ was left open. Due to this some H2O2
decomposed and the volume strength of the solution decreased. To determine the new
volume strength of the H2O2 solution, 10 mL of the solution was taken and it was diluted to
100 mL. 10mL of this diluted solution was titrated against 25 mL of 0.0245 M KMnO4 solution
under acidic condition. Calculate the volume strength of the H2O2 solution.
19. The molecular mass of an organic acid was determined by the study of its barium salt 4.290g
of salt was quantitatively converted to free acid by the reaction with 21.64 ml of 0.477 M
2+
H2SO4. The barium salt was found to have two mole of water of hydration per Ba ion and
the acid is mono basic. What is molecular weight of anhydrous acid?
20. P and Q are two elements which forms P2Q3 and PQ2. If 0.15 mole of P2Q3 weights 15.9g and
0.15 mole of PQ2 weights 9.3g, what are atomic weights of P and Q?
21. How much AgCl will be formed by adding 200 mL of 5N HCl to a solution containing 1.7g
AgNO3?
22. The reaction, 2C + O2 2CO is carried out by taking 24g of carbon and 96g O2, find out:
(a) Which reactant is left in excess?
(b) How much of it is left?
(c) How many mole of CO are formed?
(d) How many g of other reactant should be taken so that nothing is left at the end of
reaction?
23. 20 mL of a solution containing 0.2 g of impure sample of H2O2 reacts with 0.316 g of KMnO4
(acidic). Calculate:
(a) Purity of H2O2.
(b) Volume of dry O2 evolved at 27°C and 750 mm of Hg pressure.
24. Calculate the weight of lime (CaO) that can be prepared by heating 200 kg of limestone
(CaCO3) which is 95% pure.
25. A sample of hard water contains 1 mg CaCl2 and 1 mg MgCl2 per litre. Calculate the
hardness of water in terms of CaCO3 present in per 106 parts of water.
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
59
LEVEL - II
Multiple Choice Questions (Single Option Correct)
1.
3.64 gm of a metal requires 130.0 ml of 0.5M HCl to dissolve it. What is equivalent weight
of metal?
(A) 46
(B) 65
(C) 56
(D) 42
2.
The vapour density of a volatile chloride of metal is 59.5 and equivalent mass of the
metal is 24. The atomic mass of element will be …………..
(A) 96
(B) 48
(C) 24
(D) 12
3.
1g of Ca was burnt in excess of O2 and the oxide was dissolved in water to make up one
litre solution. The normality of solution is
(A) 0.04
(B) 0.4
(C) 0.05
(D) 0.5
4.
The normality of solution obtained by mixing 100 ml of 0.2M H2SO4 with 100 ml of 0.2 M
NaOH is
(A) 0.1
(B) 0.2
(C) 0.5
(D) 0.3
5.
The oxidation number of phosphorous in Ca(H2PO3)2 is
(A) +1
(B) +2
(C) +3
(D) -1
6.
In which of the following compounds, iron has lowest oxidation state?
(A) FeSO4.(NH4)2SO4.6H2O
(B) K4[Fe(CN)6]
(C) Fe0.94O
(D) Fe(CO)5
7.
In which of the following changes, there is transfer of six electrons?
(A) MnO 4 Mn2
(B) CrO24 Cr 3
(D) Cr2 O72 2Cr 3
(C) MnO 4 MnO2
8.
In which of the following phosphorus has the highest oxidation state?
(A) H4P2O7
(B) PH3
(C) H3PO2
(D) H3PO3
9.
Which of the following is not a redox reaction?
(A) BaO2 H2 SO4 BaSO 4 H2 O2
(C) 2KClO3 2KCl 3O2
(B) 2BaO O 2 2BaO2
(D) SO 2 2H2 S 2H2 O 3S
10.
Oxidation states of the metal in the minerals haematite and magnetite, respectively are
(A) II, III in haematite and III in magnetite
(B) II, III in magnetite and II in haematite
(C) II in haematite and II in magnetite
(D) III in haematite and II, III in magnetite
11.
In alkaline medium, ClO2 oxidises H2O2 to O2 and is itself reduced to Cl–. How many
moles of H2O2 are oxidized by 1 mol of ClO2.
(A) 1
(B) 3/2
(C) 5/2
(D) 7/2
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12.
A candle is burnt in a beaker until it extinguishes itself. A sample of gaseous mixture in
20
20
the beaker contains 6.08 10 molecules of N2, 0.76 10 molecule of O2 and
20
0.50 10 molecules of CO2. The total pressure is 734 mm of Hg. The partial pressure
of O2 would be
(A) 760 mm of Hg
(B) 76.0 mm of Hg
(C) 7.6 mm of Hg
(D) 0.76 mm of Hg
13.
When 2 gram of gas A is introduced an evacuated flask kept at 25 C, the pressure was
found to be 1 atmosphere if 3 g of another gas B is then added to the same flask, the
pressure becomes 1.5 atm. Assuming ideal behaviour, the ratio of molecular weights
(MA : MB) is
(A) 1 : 3
(B) 3 : 1
(C) 2 : 3
(D) 3 : 2
14.
o
2H2O2 2H2O O2 g
100 mL of x molar H2O2 gives 3 L of O2 gas under the conditions when 1 mole occupies
24 litres. The value of x is
(A) 2.5
(B) 1.0
(C) 0.5
(D) 0.25
15.
For the reaction, FeS2 Fe3 SO 2 , the n factor of FeS2 is
(A) 1
(B) 11
(C) 28
(D) 61
Multiple Choice Questions (Multiple Options Correct)
1.
Identify the correct statements among the following:
N
(A) 0.5 g-atom of oxygen contains A oxygen molecules.
4
N
(B) 0.5 g-molecule of oxygen contains A oxygen molecules.
4
(C) total electrons present in 0.5 g-atom of oxygen is 4NA.
7N
(D) total electrons present in 0.5 g-atom of O 2 is A .
4
2.
Consider the following reactions:
(i) H2 C2 O4 2NaOH Na2 C2 O4 2H2 O
(ii) H2C2O4 MnO4 H Mn2 CO2
Select the incorrect statements about these reactions
(A) Equivalent weight of oxalic acid in reaction (i) is M.
(B) Equivalent weight of oxalic acid in reaction (i) is M/2.
(C) 100 ml of 0.2 M oxalic acid are sufficient to neutralize 100 ml of 0.2 M NaOH solution.
(D) Equivalents of oxalic acid in 100 ml of 0.2 M solution of it are 40.
3.
Which of the following statements are correct for Mohr’s salt?
(A) It decolories KMnO4
(B) It is primary standard titrant
(C) It is double salt
(D) Oxidation state of Fe is +3
4.
Equal weights of X (atomic weight = 36) and Y (atomic weight = 24) are reacted to form
the compound X2Y3, which of the following is/are correct?
(A) X is the limiting reagent
(B) Y is the limiting reagent
(C) No reactant is left over
(D) Mass of X2Y3 formed is double the mass of X taken
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
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5.
In the reaction (unbalanced)
FeC2O 4 Cr2O72 Cr 3 Fe2O3 CO2
(A) n-factor of FeC2O4 is 3
(C) n-factor of Cr2O72 is 6
(B) n-factor of FeC2O4 is 2
(D) n-factor of Cr2O72 is 3
Numerical Based Single digit Integer Type
1.
What is n-factor of KMnO4, when it react with oxalic acid?
2.
The number of moles of H2 given by reaction of 48g Mg with excess of acid(HCl)
3.
0.968 g of an acid are present in 300 ml of a solution. 10 ml of this solution required
exactly 20 ml of 0.05 N KOH solution. Calculate no. of neutralisable protons. (Given mol.
weight of acid is 98)
4.
In the following reaction: xZn yHNO3 very dil aZn NO3 2 bH2 O cNH4NO3
What is the sum of the coefficients (a + b + c)?
5.
What is oxidation number of sulphur in Caro’s acid?
Numerical Based Decimal Type
1.
A and B are two elements which A2B3 & AB2 molecules. If 0.2 mole each of A2B3 and AB2
weights 17 g and 11 g respectively. The sum of atomic weight of A and B is x. The value
of x/4 is
2.
The half of equivalent weight of Na2HPO4 salt in the reaction:
2NaOH H3PO4 Na2HPO 4 2H2 O
Assertion & Reasoning Type
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation
for Statement-1.
(C) Statement-1 is True, Statement -2 is False.
(D) Statement-1 is False, Statement-2 is True.
1.
STATEMENT-1: For the reaction A x By
A y B x , (x + y = 6 and x > 0, y > 0)
two moles of B y is required for complete reaction with two moles of A x , if x and y are 4
and 2 respectively.
STATEMENT-2: In above reaction meq. of A x meq. of B x
2.
STATEMENT-1: HClO4 is stronger acid than HClO3
STATEMENT-2: Oxidation state of Cl in HClO4 is +VII and in HClO3; it is +V.
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62
Comprehension Type – I
In double titration, mainly two indicators are used, viz., phenolphthalein (HPh) and methyl orange
(MeOH) respectively. Let for complete neutralization of Na2CO3, NaHCO3 and NaOH. a, b and c
ml of standard HCl are required. The titration of the mixture may be carried by two methods as
summarized below:
Volume of HCl used with
Volume of HCl
Mixture
HPh from
MeOH from
HPh from
MeOH after first end point
beginning
beginning
beginning
NaOH + Na2CO3
(a + c)
a
a
a
c
c
(For
remaining
50%
2
NaOH
NaHCO3
Na2CO3
NaHCO3
2
2
+
c+0
(b + c)
c+0
Na2CO3)
B (For
NaHCO3)
+
a
0
2
(a + b)
a
0
2
a
2 b for remaining 50% of
100%
remaining
Na2CO3 and 100% NaHCO3
are indicated
Read the above paragraph carefully and answer the questions given below it:
N
HCl is required to react completely with 1.0 g of a sample of limestone.
10
Calculate the percentage purity of CaCO3.
(A) 65%
(B) 25%
(C) 75%
(D) 85%
1.
150 ml of
2.
A solution contained Na2CO3 and NaHCO3. 15 ml of the solution required 5 ml of N HCl
10
for neutralization using phenolphthalein as an indicator. Addition of methyl orange
required a further 15 ml of the acid for neutralization. The amount of Na2CO3 present in
the solution is
(A) 21.2 g
(B) 0.053 g
(C) 0.212 g
(D) 4.24 g
Comprehension Type – II
Read the paragraph carefully and answer the following questions:
The valency of carbon is generally 4, but its oxidation state may be –4, –2, 0, +2, +1 etc. In the
compounds containing C, H and O, the oxidation number of C is calculated as
2n nH
Oxidation number of C O
. Where nO, nH and nC are the number of oxygen, hydrogen
nC
and carbon atoms respectively.
3.
The oxidation state of C in diamond is
(A) 0
(C) –1
(B) +1
(D) +2
4.
In which of the following compounds is the oxidation state of carbon is zero.
(A) CH4
(B) CH3OH
(C) HCOOH
(D) C6H12O6
5.
In which of the following compound(s) oxidation state of C is fractional?
(A) CO
(B) CO2
(C) C3O2
(D) All
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Comprehension Type – III
Redox reactions are those in which oxidation and reduction take place simultaneously. Oxidising
agent can gain electron whereas reducing agent can lose electron easily. The oxidation state of
any element can never be in fraction. If oxidation number of any element comes out be in fraction,
it is average oxidation number of that element which is present in different oxidation states.
6.
Which of the following can be both oxidizing as well as reducing agent?
(A) H2
(B) I2
(C) H2O2
(D) All of these
7.
The oxidation number of sulphur in K2S2O8 is
(A) +2
(B) +4
(C) +7
(D) +6
Match the Following
1.
Match the underlined atoms in compounds given in Column-I with their corresponding
correct oxidation states given in Column-II:
Column- I
Column-II
(A)
CaOCl2
(P)
+1
(B)
(Q)
0
N3
(C)
H2SO5
(R)
2
(D)
(S)
Cl CCl2 C C CHO
1
(T)
+3
2.
Match the following List - I with List - II
List – I (Redox reaction)
(P) K 4 Fe CN
Fe3+ + CO2 +NO3
6
(Q) FeS 2
Fe3+ + SO2
List – II (n-factor of reactant)
(1) 61
(2)
11
(R)
3Br2 + 6NaOH
NaBrO3 + 5NaBr + 3H2O
(3)
5/3
(S)
As 2 S3
As+5 + SO 42-
(4)
28
Codes:
(A)
(B)
(C)
(D)
3.
P
1
1
3
4
Q
2
3
1
3
R
3
2
4
2
S
4
4
2
1
Match the following List I (Reaction)with List - II(Eq. mass of reactant).
List-I
List-II
P
1
M.wt./20
NH3
NO3
Q
Fe2S 3 FeSO 4 SO2
2
M.wt./2
R
CaCO3 2HCl CaCl2 H2O CO 2
3
M.wt./8
S
CuS
CuSO 4
4
50
The correct option is
(A) P → 3; Q → 1; R → 2, 4; S → 3
(B) P → 4; Q → 3, 4; R → 1, 4; S → 1, 2
(C) P → 1, 2, 3; Q → 2, 4; R → 1; S → 2, 3
(D) P → 3; Q → 2; R → 1, 2; S → 4
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
64
4.
1
2.
Answer questions 1, 2 and 3 by appropriately matching the information given in the three
columns of the following table.
Column – 1
Column – 2
Column – 3
(I)
88g CO2
(i)
0.25 mole of molecules
(P)
3.613 1024 atom
(II)
6.022 1023
molecules of H2O
(ii)
2 mole of molecules
(Q)
3.011 1023 atom
(III)
5.6 of O2 at STP
(iii)
1 mole of molecules
(R)
2.7099 1024 atom
(IV)
72g of O3
(iv)
1.5 mole of molecules
(S)
1.8066 1024 atom
The only correct combination is
[A] (IV) (iii) (R)
[C] (III) (iii) (P)
[B] (I) (ii) (Q)
[D] (I) (ii) (P)
Which one is only incorrect combination is
[A] (I) (iii) (Q)
[C] (III) (i) (Q)
[B] (I) (ii) (P)
[D] (IV) (iv) (R)
3
The only correct combination of 6.022 1023 molecules of H2O is
[A] (III) (iii) (P)
[B] (II) (iii) (S)
[C] (IV) (iii) (R)
[D] (I) (ii) (Q)
5.
Answer the following by appropriately matching the lists based on the information given in
the paragraph
Match the following List- I with List - II.
List – II
List – II
(I) P2H4 PH3 P4H2
(P) E
3M
4
(II) I2 I IO3
(Q) E
3M
5
(III) MnO 4 Mn2 H2O Mn3O 4 H
(R) E
15M
26
(IV) H3PO2 PH3 H3PO3
(S) E
5M
6
(T) E
M
28
(U) E
M
2
1.
Which of the following options has correct combination considering List I with List II?
(A) (I) (S)
(B) (II) (P)
(C) (I) (T)
(D) (II) U
2.
Which of the following options has correct combination considering List I with List II?
(A) (III) (P)
(B) (IV) (Q)
(C) (III) (R)
(D) (IV) S
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65
LEVEL - III
Multiple Choice Questions (Single Option Correct)
1.
An aqueous solution of 6.3 g of oxalic acid dihydrate is made up to 250 ml. The volume of
0.1 N NaOH required to completely neutralize 10 ml of this solution is:
(A) 40 ml
(B) 20 ml
(C) 10 ml
(D) 4 ml
2.
The empirical formula of an organic compound containing carbon and hydrogen is CH2.
The mass of one litre of this organic gas is exactly equal to that of one litre of N2.
Therefore, the molecular formula of the organic gas is:
(A) C2H4
(B) C3H6
(C) C6H12
(D) C4H8
3.
A compound contains atoms X, Y, Z. The oxidation number of X is +2, Y is +5 and Z is
–2. The possible formula of compound is
(A) XYZ2
(B) Y2(XZ3)2
(C) X3(YZ4)2
(D) X3(Y4Z)2
4.
19.7 kg of gold was recovered from a smuggler. How many atoms of gold were recovered
(Au = 197)?
(A) 100
(B) 6.02 × 1023
24
(C) 6.02 × 10
(D) 6.02 × 1025
5.
The specific heat of a metal is 0.11 cal/g oC and its equivalent weight is 18.61. Its exact
atomic weight is:
(A) 58.2
(B) 29.1
(C) 55.83
(D) 27.91
6.
Among the following, identify the species with an atom in +6 oxidation state:
(A) MnO 4
(B) Cr2O3
(C) HClO 4
(D) Cr2 O72
7.
Equivalent weight of oxidizing agent in the reaction
SO 2 2H2S 3S 2H2O, is
(A) 32
(B) 64
(C) 16
(D) 8
8.
In which of the following pairs, there is greatest difference in the oxidation number of
underlined elements?
(A) NO2 and N2O4
(B) P2O5 and P4O10
(C) NO2 and N2O3
(D) SO2 and SO3
9.
10 gram of a piece of marble was put into excess of dilute HCl acid. When the reaction
was complete, 1120 cm3 of CO2 was obtained at STP. The percentage of CaCO3 in the
marble is
(A) 25%
(B) 50%
(C) 75%
(D) 10%
10.
A solution of 10 mL
M
FeSO4 was titrated with KMnO4 solution in acidic medium. The
10
amount of KMnO4 solution will be used
(A) 5 mL of 0.1 M
(B) 10 mL of 0.1 M
(C) 10 mL of 0.5 M
(D) 10 mL of 0.02 M
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
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Numerical Based Single digit Integer Type
1.
When the following redox reaction is balanced using simplest “natural numbers”, the
stoichiometric coefficient of H2O is X. Calculate X/2?
Cr OH3 IO
I CrO42 H2O
2 OH
2.
KMnO4 oxidises oxalic acid in acidic medium. The number of CO2 molecules produced as
per the balanced equation is X. Calculate X/10?
Numerical Based Decimal Type
1.
Number of moles of NH3 formed when 0.535 g of NH4Cl is completely decomposed by
NaOH is
NH4 Cl NaOH NH3 NaCl H2 O
2.
Two solution of a substance (non-electrolyte) are mixed in following manner
480 mL of 1.5 M
+
520 mL of 1.2 M
(solution I)
(solution II)
Molarity of final solution is
3.
1.0 g of metal nitrate gave 0.86 g of metal sulphate. Calculate equivalent weight of metal.
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
67
ANSWERS TO CHAPTER PRACTICE PROBLEMS (CPP)
Subjective:
LEVEL – I
1.
Mn+7 + 5e Mn2+ (acidic medium)
P4O6 + KMnO4 H3PO4 + MnCl2
Meq of KMnO4 = Meq of P4O6
Meq of KMnO4 = 10 0.2 1
or, 20 5 × X = 10 0.2 1
or, X = 0.02 M
W
8 1000 20 0.02 5
220
W = 0.055 gm
% purity of P4O6 = 55%.
2.
20 NHIO3 = 35.5 0.16
35.5 0.16
0.284 (N)
20
N factor of HIO3 = 5
176
Mass of HIO3 0.284
= 10 gm/lt.
5
NHIO3
3.
Cr2 O72 14H 6I
2Cr 3 7H2 O 3I2
1
1
S2 O32 I2 S4 O62 I
2
2
Equivalent weight of K2Cr2O7 =
mol. wt.
294
49
changeinONper mole
6
milli equivalent of 10 mL of K2Cr2O7 solution = milli equivalent of iodine
= milli equivalent of sodium thiosulphate =
1
16 0.64
25
weight per 10 mL = 0.64 × 49= 31.36 mg.
concentration of K2Cr2O7 in milligram per litre = 31.36 × 100 = 3136 mg/L.
LEVEL – II
4.
Meq of
Meq of
Meq of
Meq of
HCl = 50 × 0.1 = 5
NaOH = 10 × 0.16 = 1.6
HCl = Meq of Na2CO3 + Meq of NaOH
Na2CO3 in 50 ml solution = 5 – 1.6 = 3.4
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
68
In 250 ml Meq of Na2CO3 = (3.4 × 250) / 50 = 17
W
2 1000 17
106
W = 0.901 gm
% purity =
0.901
100 90.1 ~
90%.
1
(NH4 )2 SO 4
NaOH
1.5 g NH3
Impurity
5.
100 mL NH2SO4
160 mL N/2 NaOH
160 mL N/2 NaOH = 80 mL N NaOH 80 mL NH2SO4
Consumed acid = 100 – 80 = 20 mL N H2SO4
2NH3 H2SO 4 (NH4 )SO 4
2000 mL N
2 17
(NH4 )2 SO4 2NaOH Na2 SO4 2NH3 2H2O
132
2 17
2000 mL N H2SO4 2 × 17 g NH3 132 g (NH4)2SO4
132 20
20 mL NH2SO4
= 1.32 g (NH4)2SO4
2000
1.32
% (NH4)2SO4 =
100 = 88.
1.5
Objective:
1.
A
2.
B
3.
D
4.
A
5.
A
6.
A, B
7.
A, B, D
8.
A, B
9.
1
10.
5
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69
ANSWERS TO ASSIGNMENT PROBLEMS
Subjective:
LEVEL - I
oxygen
6 10 3
106
106 = 6 ppm
solution
103
wt. of
1. ppm of oxygen =
wt. of
2. (b) Molarity – It will decrease with the increase in temperature. (because of increase in
volume)
(a) Molality – It does not change with temperature
3. 1 Molar solution means one mol solute present in 1000 ml of solution while one molal
solution means 1 mol of solute is present in one kilogram (1000 ml, if density is one) of
solvent. It means in one molar solution the solvent is always less than 1000 ml solution.
Therefore the one molar solution is more concentrated then one molal solution.
4. Molarity of a solution is defined as the number of moles of solute present in one litre of
solution while the molality is the number of moles of solute in one kilogram of solvent.
Molarity depends upon the volume of the solution which changes with temperature; hence
molarity depends upon the temperature. On the other hand molality depends upon the mass
of solvent, which does not vary with temperature hence molality does not depend upon
temperature. Therefore, when there is a chance of changing the temperature during the
experiment, molality is preferred over molarity.
5. Molality =
wt. of glucos e
1000
mol. wt. of glucose wt. of solvent
2.82 1000
0.52 m
180
30
No. of moles of glucose =
No. of moles of H2 O
wt. of glucose
2.82
0.016
Mol. wt. of glucose 180
Moles of
mol.wt of
water 30
1.67
water 18
Mole fraction of
glucose (XGlu )
Moles.
Moles of glucos e
of water Moles of
Mole fraction of water XH2 O
Glucose
0.016
0.01
0.016 1.67
1.67
0.99
0.016 1.67
6. (i) 86% H2 SO4 means 86 gms of sulphuric acid is present in 100 g of the solution
Volume of the 100 g of acid =
wt. of
Mol. Mass
(ii) M1V1 M2 V2
Molarity =
Mass
100g
= 55.9 ml
Density 1.787g / ml
subs tance
1000
86 1000
=
15.7M
of subs tance Vol. of solution 98 55.9
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70
M1 15.7, M2 0.2, V2 1000
ml, V1 ?
0.2 1000
12.74ml
15.7
(iii) Mass of the solution = 100 g
Mass of the solute = 86 g
Mass of the solvent = 100 – 86 = 14 g
wt. of subs tance
1000
86 1000
Molality
Mol. wt. of subs tance wt. of solvent(gm) 96
14
V1
62.86 m
7. 1 M solution of NaNO3 means, 1 mol of NaNO3 is present in 1 lit of solution.
1 mol of NaNO3 = 23+14+48 = 85 g
Mass of solution = volume of solution density
= 1000 1.25 1250 gm
85 g of NaNO3 (solute) is present in 1250 gm of solution
wt of solvent = wt. Of solution – wt. of solute
= 1250 – 85 = 1165 gm
100
1000
Molality = moles of solute
= 1
0.86 m
wt. of solvent
1165
8. Molar mass of Ba(OH)2 .8H2O =137+(2 17)+( 8 18) = 315 g mol-1
WB
1000
MB Wt of solvent
Where WB wt. of substnace 5.6g
Molality of Ba(OH)2 m
MB mol. Mass of substance = 315 g
wt of solvent = 100 g
5.6 1000
Hence, m =
0.178m
315 100
Ba(OH)2 will dissociate to followingequation
Ba(OH)2
Ba 2OH
1mol
1mol
2mol
Hence, the molality of OH ion = 2 molality of Ba(OH)2 2 0.178 0.356g
9. 13% H2 SO4 solution (by weight) means that 13 g H2 SO4 is present in 100 g of solution.
Molarity =
WB
1000
MB Volume of solution
Volume of solution =
Mass of the solution
100
98c.c
Density of the solution
1.02
13 1000
1.35 mole / litre
98
98
Mass of solvent = Mass of solution – Mass of solute = 100 – 13= 87 g
W
1000
Molality= B
MB Massof solvent
Hence, molarity =
=
13 1000
1.52mol / Kg
98
87
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71
10. (i)
5H2 S 2MnO 4 16H
5S 2Mn2 8H2O
(ii) 12OH 6Br2
10Br 2BrO3 6H2 O
11. KNO3 H2 SO 4
HNO3 KHSO4
1 mole KNO3 gives 1 mole of HNO3
85
50
100 moles of HNO
moles of KNO3 =
3
101
50 85 63
WHNO3
26.51 gm
100 101
3
O2
2
1 mole of KNO3 gives 3/2 moles of oxygen
moles of KClO3 = 0.25
3
moles of O2 = 0.25
2
3
volume of O2 at STP = 0.25 22.4 8.4 litre
2
12. KClO3
KCl
13. 58.7 gm Ni required = 4 28 gm of CO
4 28
1 gm Ni required =
= 1.908 gm of CO
58.7
14. CuSO4.5H2O 1.5 litre, 0.01 M solution
moles of CuSO4 required = 1.5 0.01 = 0.015
W CuSO4 .5H2 O required = 0.015 249.5
= 3.7425 gm
15. Number of gm equivalent of Na2S2O3 = number of equivalent of K2Cr2O7
0.1 15 = 10 N
N = 0.15
16. MnO2 + 4 HCl
MnCl2 + 2H2O + Cl2
Number of moles of Cl2 =
1.78
0.07946
22.4
Number of moles of MnO2 = 0.07946
Mass of MnO2 = 0.0794 87 = 6.913 g
Number of moles of HCl = 4 0.07946 = 0.3178
mass of HCl = 0.3176 36.5
Let the volume of HCl = V ml
V 1.2
4
4 0.317 36.5 = 241.66 ml
100
2+
17. MnO4 would convert to Mn . Therefore its ‘n’ factor would be 5.
-3
-3
Equivalents of MnO4 = 1.61 10 5 = 8.05 10
n+
-3
Equivalents of A = 8.05 10
’n’ factor of AO3 = 5 – n
(5 – n) 2.68 10-3 = 8.05 10-3
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
72
18. If a solution of H2O2 has normality = N, it means that 1 litre of the solution contains N
equivalents.
N
equivalents.
1000
1
N
moles of H2O2 in 1 ml
=
2 1000
1 1
N
Moles of O2 it gives =
2 2 1000
1 1
N
Volume of O2 at STP in mL = 224400
2 2 1000
1 mL of it would contain
= 5.6 N
Volume of strength = 5.6 Normality
Normality of H2O2 solution =
0.0245 25 5
= 0.306
10
Normality of the 100 mL H2O2 solution = 0.306
(that has been diluted)
Normality of the original 10 mL H2O2 solution = 0.306 10 = 3.06
Volume strength = 5.6 3.06 = 17.136
19. Meq. of barium salt = Meq. of acid
4.290
1000 = 21.64 0.4777 2
M/ 2
Molecular weight of salt = 415.61
415.61 137 36
Molecular weight of anion =
121.31
2
Molecular weight of acid = 121.31 + 1 = 122.31
20. Let atomic weight of P and Q are a and b respectively
Molecular weight of P2Q3 = 2a + 3b and Molecular weight of PQ2 = a + 2b
Now given that 0.15 mole of P2Q3 weigh 15.9g
15.9
wt.
(2a 3b)
mole
0.15 mol. wt.
9.3
0.15
Solving these two equations
b = 18, a = 26
Similarly, (a 2b)
21.
AgNO3
Meq. mixed
+
1.7
1000
170
= 10
Meq. after reaction 0
Meq. of AgCl formed = 10
HCl AgCl + HNO3
200 5
= 1000
990
0
10
0
10
w
1000 10
143.5
wAgCl = 1.435 g
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
73
22.
2C + O2 2CO
Mole before reaction
24
12
96
32
=2
=3
0
Mole after reaction
0
2
2
mole ratio of C : O2 : CO : : 2 : 1 : 2
(a) O2 is left in excess.
(b) 2 mole of O2 or 64 g O2 is left
(c) 2 mole of CO or 56 g CO is formed
(d) To use O2 completely total 6 mole of carbon or 72g carbon is needed.
23. Redox changes are:
+7
+2
5 e + Mn Mn
O 21 O02 2e
(a) Meq. of H2O2 = Meq. of KMnO4
w 1000 0.316
1000
34 / 2
M /5
w 2 1000 0.316 5 1000
34
158
w H2 O2 0.17g
0.2 g impure sample of H2O2 has 0.17 g pure H2O2
0.17 100
% of H2O2 =
85%
0.2
(b) Now Eq. of O2 = Eq. of KMnO4
w
0.316 5
32 / 2
158
w O2 0.16g
750
0.16
V
0.0821 300 VO2 124.79mL
760
32
24. CaCO3 CaO + CO2
100 gm
56 gm
100 gm contains 56 gm
(Pure CaCO3 = 200
190 1000 =
95
190 kg 190 1000 gm)
100
56
190 1000 106.4 kg
100
25. CaCl2 CaCO3
111 mg
100mg
100
1 mg
= 0.90 mg
111
MgCl2
CaCO3
95 mg
100 mg
100
1 mg
= 1.05 mg
95
Total mass of CaCO3 = 0.90 + 1.05 = 1.95 mg/litre = 1.95 ppm
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
74
LEVEL - II
Multiple Choice Questions (Single Option Correct)
1.
5.
9.
13.
C
C
A
A
2.
6.
10.
14.
B
D
D
A
3.
7.
11.
15.
C
D
C
B
4.
8.
12.
A
A
B
A, B, C
4.
C, D
3
4.
8
4.
D
Multiple Choice Questions (Multiple Options Correct)
1.
A, C
5.
A, C
2.
A, C
3.
Numerical Based Type
1.
5.
5
6
2.
2
3.
Numerical Decimal Based Type
1.
7.5
2.
35.5
Assertion & Reasoning Type
1.
C
2.
A
Comprehension Type
1.
5.
C
C
2.
6.
B
C
3.
7.
A
D
Match the Following
1.
(A P, S); (B Q, S); (C R, S); (D P, Q, T)
2.
A
3.
A
4.
1.
D
2.
A
5.
1.
A
2.
C
3.
B
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
75
LEVEL - III
Multiple Choice Questions (Single Option Correct)
1.
5.
9.
A
C
B
2.
6.
10.
A
D
D
3.
7.
C
C
4.
8.
D
D
Numerical Based Single digit Integer Type
1.
5
2.
1
Numerical Based Decimal Type
1.
0.01
2.
1.344
3.
38
P-2123-CBSE-P1-CHEMISTRY- REDOX REACTION & STOICHIOMETRY
