Chapter 7 Preparation
CHAPTER 7
ASSESSMENT
Identifying Reaction Types
BLM 7.0.1
(continued)
9. Classify each of these reactions into one of the following reaction types:
formation (F),single replacement (SR), simple decomposition (SD), double replacement
(DR), combustion (C), or other (O).
F
(a) Ti(s) + 2Cl2(g) → TiCl4(ℓ)
SD (c) HCOOH(ℓ) → C(s) + H2(g) + O2(g)
C
(e) CH3COOH(ℓ) + 2O2(g) → 2CO2(g) + 2H2O(g)
F
(g) 2C(s) + 4H2(g) + O2(g) → 2CH3OH(ℓ)
SR (i) Br2(ℓ) + 2NaI(aq) → 2NaBr(aq) + I2(s)
SD (k) 8H2SO4(ℓ) → 8H2(g) + S8(s) + 16O2(g)
SR (m) Sn(NO3)2 (aq) + Cd(s) → Sn(s) + Cd(NO3)2(aq)
F
(o) 2Ag(s) + O2(g) + H2(g) → 2AgOH(s)
DR (q) H3PO4(aq) + 3NaOH(aq) → 3HOH(ℓ) + Na3PO4(aq)
DR (s) Ba(NO3)2(aq) + 2NaCl(aq) → BaCl2(s) + 2NaNO3(aq)
SR (u) I2(s) + Na2Se(aq) → 2NaI(aq) + Se(s)
F
(w) 2Au(s) + 3Cl2(g) → 2AuCl3(s)
C
(y) 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
1 6H10O5(s) + ____O
6 2(g) → ____CO
6
5
(a)____C
2(g) + ____H
2O(g)
1
3
2
(c)____Sc
2O3(s) + ____H2O(ℓ) → ____Sc(OH)3(s)
BLM
7.0.3
1 CH3COOH(ℓ) + ____O
2
2
2
(e)____
2(g) → ____CO
2(g) + ____H
2O(g)
1
5 2(g) → ____CO
3
4 2O(g)
(g)____C
3H8(g) + ____O
2(g) + ____H
2
2
3
2
(i)____Ca(s)
+ ____C(s)
+ ____O
2(g) → ____CaCO3(s)
6 2Cl2(ℓ) + ____NH
16
1
1
12
(k)____S
3(g) → ____S4N4(s) + ____S8(s) + ____NH4Cl(s)
2
1
1
2
(m)____Ag(s)
+ ____O
2(g) + ____H2(g) → ____AgOH(s)
1
1
1
1
(o)____CrCl
→ ____MgCl
2(aq) + ____Mg(s)
2(aq) + _____Cr(s)
2 8H18(ℓ) + ____O
25 2(g) → ____CO
16
18
(q)____C
2(g) + ____H
2O(g)
8
1 8(s) + ____O
12 2(g) → ____CuSO
8
(s)____Cu(s)
+ ____S
3(s)
8
8
1 8(s) + ____O
16 2(g)
(u)____CuSO
+ ____S
4(s) → ____Cu(s)
8
1
16 2(g) → ____PbSO
8
(w)____Pb(s)
+ ____S
8(s) + ____O
4(s)
1
3
1
3
(y)____BCl
(g) + ____H
O(ℓ) → ____H
BO (s) + ____HCl(g)
1. Cyclopentane, C5H10(ℓ), reacts with oxygen to produce carbon dioxide
and water vapour.
BLM 7.04
Supply
formulas
and
balance
2 C5H10(l) + 15 O2(g)  10 CO2(g) + 10 H2O(g)
3.Phosphoric acid is neutralized by cesium hydroxide solution to produce
cesium phosphate solution and liquid water.
1 H3PO4(aq) + 3 CsOH(aq)  3 HOH(l) + 1 Cs3PO4(aq)
Note: HOH(l) = H2O(l)
5. Lead(II) acetate solution reacts with zinc metal to form zinc acetate
solution and solid lead.
1 Pb(CH3COO)2(aq) + 1 Zn(s)  1 Zn(CH3COO)2(aq) + 1 Pb(s)
7. Calcium chlorate solution reacts with lithium sulfate and calcium sulfate
solid forms in a lithium chlorate solution.
1 Ca(ClO3)2(aq) + 1 Li2SO4(aq)  1 CaSO4(aq) + 2 LiClO3(aq)
9.Octane (C8H18(ℓ)) reacts with oxygen gas to produce carbon dioxide and
liquid water in a bomb calorimeter.
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(l)
1. Ethylene (ethene), C2H2(g), forms from its elements.
(F)
2 C(s) + 1 H2(g)  1 C2H2(g)
2. Dihydrogen dioxide (hydrogen peroxide), H2O2(ℓ), decomposes into its elements.
(SD)
1 H2O2(l)  1 H2(g) + 1 O2(g)
BLM 7.0.5
3. Methanal, CH2O(ℓ), burns in oxygen gas.
Predictions:
C, SD, F
(C)
1 CH2O(l) + 1 O2(g)  1 CO2(g) + 1 H2O(g)
5. Cyclohexane, C6H12(ℓ), undergoes combustion. (C)
Review info
about states
on WS
1 C6H12(l) + 9 O2(g)  6 CO2(g) + 6 H2O(g)
7. Benzene, C6H6(ℓ), combusts. (C)
2 C6H6(l) + 15 O2(g)  12 CO2(g) + 6 H2O(g)
Hint:
determine
reaction
type 1st
9.
1 P4(s)
6 Cl2(g) → ____
4 PCl (g)
________
+ ________
3
(F)
1
9 O2(g) → 7 CO2(g) + 4 H2O(g)
11. ____C
7H8(ℓ) + ________
(C)
1. Liquid bromine reacts with a sodium iodide solution. (SR)
BLM 7.0.6
Predictions:
SR and DR
Review info
about states
on WS
1 Br2(l) + 2 NaI(aq)  2 NaBr(aq) + 1 I2(s) or (aq)
3. Calcium chloride solution reacts with ammonium sulphide (DR)
solution.
1 CaCl2(aq) + 1 (NH4)2S (aq)  1 CaS(s) + 2 NH4Cl(aq)
5. Ethanoic acid reacts with a solution of potassium hydroxide. (DR)
1 CH3COOH(aq) + 1 KOH(aq)  1 KCH3COO(aq) + 1 HOH(l)
7. Magnesium metal is put into a silver nitrate solution. (SR)
1 Mg(s) + 2 AgNO3(aq)  1 Mg(NO3)2(aq) + 2 Ag(s)
Determine
reaction
type 1st
9. Silver nitrate solution is mixed with a lithium chloride solution. (DR)
1 AgNO3(aq) + 1 LiCl(aq)  1 AgCl(s) + 1 LiNO3(aq)
11. Aluminium metal is added to a chromium(II) sulfate solution. (SR)
2 Al(s) + 3 CrSO4(aq)  1 Al2(SO4)3(aq) + 3 Cr(s)
13. Aqueous chlorine is mixed with a solution of sodium selenide. (SR)
2 Cl2(aq) + 2 Na2Se(aq)  4 NaCl(aq) + 1 Se2(s)
Chemistry 20
Chapter 7
PowerPoint presentation by
R. Schultz
robert.schultz@ei.educ.ab.ca
7.1 Reactions in Aqueous
Solution
• Solution reaction equations are often
written as net-ionic equations
• Consider reaction of lead (II) nitrate
solution with potassium iodide solution:
1 Pb(NO3)2(aq) + 2 KI(aq)
1 PbI2(s) + 2 KNO3(aq)
equation called complete balanced equation or
formula equation
Recall from previous unit that electrolytes
dissociate (or ionize) in water
7.1 Reactions in Aqueous
Solution
• leads to the ionic equation or total ionic
equation
• dissociate all aqueous electrolytes and
write all strong acids in ionic form
1 Pb2+(aq) + 2 NO3‾(aq) + 2 K+(aq) + 2 I‾(aq)
1 PbI2(s) + 2 K+(aq) + 2 NO3‾(aq)
note that numbers from balanced equation are part of ionic equation
• now look for spectators, chemical entities
that don’t change
• cancel them out
• this leads to net ionic equation………
7.1 Reactions in Aqueous
Solution
• net ionic equation:
1 Pb2+(aq) + 2 NO3‾(aq) + 2 K+(aq) + 2 I‾(aq)
1 Pb2+(aq) + 2 I‾(aq)
1 PbI2(s) + 2 K+(aq) + 2 NO3‾(aq)
1 PbI2(s)
all precipitation reactions will have this
basic type of net-ionic equation
try Practice Problem 2a, page 264: write ionic and netionic equations – note: original equation is unbalanced
Practice
Problem 2a,
page 264
7.1 Reactions in Aqueous
Solution
• balanced:
3 Ba(ClO3)2(aq) + 2 Na3PO4(aq)
1 Ba3(PO4)2(s) + 6 NaClO3(aq)
• ionic:
3 Ba2+(aq) + 6 ClO3‾(aq) + 6 Na+(aq) + 2 PO43‾(aq)
1 Ba3(PO4)2(s) + 6 Na+(aq) + 6 ClO3‾(aq)
• net ionic:
3 Ba2+(aq) + 2 PO43‾(aq)
1 Ba3(PO4)2(s)
7.1 Reactions in Aqueous
Solution
• Single replacement example:
• complete balanced equation:
1 Zn(s) + 1 Cu(NO3)2(aq)
1 Zn(NO3)2(aq) + 1 Cu(s)
• ionic equation:
1 Zn(s) + 1 Cu2+(aq) + 2 NO3‾(aq)
1 Zn2+(aq) + 2 NO3‾(aq) + 1 Cu(s)
• net-ionic equation:
1 Zn(s) + 1 Cu2+(aq)
1 Zn2+(aq) + 1 Cu(s)
metal ions and metal elements are different!
Try Practice Problem 1c, page 264 (equation is balanced)
7.1 Reactions in Aqueous
Solution
• ionic
2 Al(s) + 3 Cu2+(aq) + 6 Cl‾(aq)
3 Cu(s) + 2 Al3+(aq) + 6 Cl‾(aq)
• net-ionic
2 Al(s) + 3 Cu2+(aq)
3 Cu(s) + 2 Al3+(aq)
all metal with ionic solution reactions will be similar
do worksheet BLM 7.1.2 questions 1-7
note that questions 3 and 7 are special –
talk to me
7.1 Reactions in Aqueous
Solution
• Techniques of Qualitative Analysis
what is
present,
nothing about
amount –
concentration,
mass, etc
– solution colour – see chart page 11 of Data
Booklet
– flame colour – see chart page 6 of Data
Booklet: I don’t like chart titles – metals
themselves don’t give colours – metal ions do
7.1 Reactions in Aqueous
Solution
selective precipitation –
solutions are added to the
unknown solution to see
whether or not precipitates
form
solubility chart page 6 Data Booklet
also need flame test colour chart,
page 6 and ion colour chart, page 11
Thought Lab 7.1 page 267
1. Red flame and ppt with
OH‾(aq): Ca2+(aq)
2. Red flame and ppt with
SO42‾(aq): Ca2+(aq), Sr2+(aq)
3. If all traces of Ca2+(aq) and
Sr2+(aq) are removed,
flame colour?
yellow because NaOH(aq)
and Na2SO4(aq) have been
added
7.1 Reactions in Aqueous
Solution
• Thought Lab 7.1, question 4
a) colourless solution; no ppts
• b) blue solution; ppt with e.g.
no ppt with e.g.
• c) colourless solution
ppt with e.g.
no ppt with e.g.
• d) blue solution
ppt with e.g.
no ppt with e.g.
Worksheet BLM 7.1.5
7.2 Stoichiometry and
Quantitative Analysis
• Stoichiometry – a method of predicting
quantities of reactants or products of a
chemical reaction from quantities of
another substance in the reaction
concept is based on mole
ratios just as # sandwiches
is based on #’s of each type
of piece
to double # sandwiches,
double # of each
sandwich piece
figure 7.7,
page 271
7.2 Stoichiometry and
Quantitative Analysis
based on mass
• Type 1: Gravimetric Stoichiometry
• Example: Practice Problem 12, page 278
2 Mg(s) + 1 O2(g)
n
m
looking for
given
n1
4.86 g
xs
2 MgO(s)
n2
?
4.86 g
 0.200 mol
g
24.31 mol
2
n2  0.200 mol   0.200 mol
2
n1 
?  m  0.200 mol  40.31g mol  8.06 g
7.2 Stoichiometry and
Quantitative Analysis
• Example: Practice Problem 15, page 278
• Unbalanced equation given – balance it
4 PCl3(l)
• 1 P4(s) + 6 Cl2(g)
n
m
not part of
molar mass
looking for
given
n1
323 g
n1 
323 g
 4.56 mol
70.90 g mol
n2  4.56 mol 
4
 3.04 mol
6
?  m  3.04 mol  137.32 g mol  417 g
Do Worksheet 30,
Investigation 7.B.1
n2
?
based on
solution
concentrations
and volumes,
but same
general
strategy
7.2 Stoichiometry and
Quantitative Analysis
• Type 2: Solution Stoichiometry
• Example: Practice Problem 16, page 282
1 MgCl2(aq) + 2 AgNO3(aq)
n2
0.50 mol/L
V=?
n1
0.30 mol/L
60 mL
2 AgCl(s) + 1 Mg(NO3)2(aq)
n1  c V  0.30 mol L  0.060 L  0.018 mol
looking for
given
1
n2  0.018 mol   0.0090 mol
2
V
0.0090mol
 0.018 L or 18 mL
mol
0.50 L
7.2 Stoichiometry and
Quantitative Analysis
• Note that in both types of stoichiometry:
– step 1 is to write a balanced equation
– step 2 is to convert the given quantity into
moles
– step 3 is to convert the moles of the given
quantity into moles of the quantity you are
looking for
– step 4 is to convert the moles of the quantity
you are looking for into the needed variable
this is true in all types of stoichiometry!
7.2 Stoichiometry and
Quantitative Analysis
• Type 3: Gas Stoichiometry
• You will learn how to do this in our Gases
Unit (next unit)
7.2 Stoichiometry and
Quantitative Analysis