Stoichiometry: Calculations with Chemical Formulas and Equations
(Chapter 3) Past Quiz and Test Questions – Answer Key
My answers are in bold faced italics and underlined – to this point I have only included
answers up to question 9.
1.
Balance the following equations
Al2(SO4)3 (aq) +
3 Ba(NO3)2 (aq) ----> 2 Al(NO3)3 (aq) + 3 BaSO4 (s)
C3H8O2 (R) + 4 O2 (g)
--------> 3 CO2 (g) +
4 H2O (g)
P2O5 (g) + 3 H2O (R) ---->2 H3PO4 (aq)
B10H18 + 12 O2
-----> 5B2O3
N2(g) + 3 H2 (g) ----->
2.
+
9 H2O
2 NH3 (g)
a.How many grams of Cl are in 113.5-g of C2Cl6?
?gCl =
113.5gC Cl
2
1
b.
213gCl
= 102.0gCl
237gC2Cl6
45.0gC H O
12
22
11
x
1
11x6.022x1023 atomsO
= 8.72x1023 atomsO
342gC12 H22O11
How many moles of CaBr2 are required to obtain 4.5 x 1021 atoms of Br?
4.5x1021 atomsBr
?molCaBr =
2
d.
x
How many atoms of O are in 45.0-g of C12H22O11?
? atomsO =
c.
6
1molCaBr
x
= 3.7x10
2
−
3
molCaBr
23
2x6.022x10 atomsBr
1
How many grams of C are required to make 1.35-mol of CH4?
?gC =
1.35molCH
4
1
x
12gC
= 16.2gC
1molCH4
e.How many moles of NH3 are contained in 250.0-g of NH3?
?molNH3 =
250.0gNH 1molNH
3x
3 = 14.7molNH3
1
17gNH3
2
3.
Given:
Consider the following chemical equation.
20.0 g K2CO3
30.0 g AlBr3
3 K2CO3 (aq) + 2 AlBr3 (aq)
Stoichiometry:
3 mol K2CO3
3 x 138 g K2CO3
--------> 6 KBr (aq) + Al2(CO3)3 (s)
2 mol AlBr3
6 mol KBr
2 x 267 g AlBr3
6 x 119 g KBr
1 mol Al2(CO3)3
1 x 234 g Al2(CO3)3
20.0-g of K2CO3 are reacted with 30.0-g of AlBr3.
a.
b.
Balance the equation.
Which of the reactants is the limiting reactant?
20.0gK2CO3 =
Compare
0.0483 3x138gK2CO3
30.0gAlBr
3
= 0.0562
2x267gAlBr3
Since the stoichiometric ratio involving K2CO3 is less than that involving AlBr3,
K2CO3 is the limiting reactant.
c.
How many grams of Al2(CO3)3 can be formed?
?gAl2 (CO3 )3 =
20.0gK CO 1x234gAl (CO )
2
3x
2
3 3 = 11.3gAl2 (CO3 )
13x138gK2CO3
Note this calculation is based on the limiting reactant, not the excess.
d.
Suppose the reaction is carried out and 7.5-g of Al2(CO3)3 are formed. What is the
percent yield?
7.5gAl2 (CO3 )3 Produced
% yield = 11.3gAl2 (CO3 )3 Calculated x100% = 66.4% yield
e.
How much of the excess reactant remains after the reaction if all of the limiting
reactant is consumed?
First, figure out how much of the excess reactant, AlBr3, was used.
?gAlBr3used =
20.0gK CO 2x267gAlBr
2
3x
3 = 25.8gAlBr3used
13x138gK2CO3
Excess AlBr3 = 30.0 g – 25.8 g = 4.2 g of excess reactant left over after the reaction.
4.
Determine the empirical and molecular formulas for epinephrine, a hormone secreted
into the blood stream during times of danger or stress. Its elemental composition is
59.0% C, 7.1% H, 26.2% O, and 7.7% N. Its molecular weight is 183.
C:
59.0 = 4.92
12
H:
O:
N:
4.
92
0.55 = 8.95
Divide by smallest(0.55) :
7.1
7.1
= 7.1
1
= 12.91
0.55
1.6375 = 2.98
0.55
26.2 = 1.6375
16
0.55 = 1
0.55
7.7
= 0.55
14
Since these are all close to an integer, round to the nearest integer for a ratio of: 9:13:3:1 or
C9H13O3N for the empirical formula. Since the formula weight of the empirical formula is
183 and matches the molecular weight, the molecular formula is also C9H13O3N.
5.
Balance each of the following equations.
C12H22O11 (s) + 12 O2
Mg (s) +
2HCl (aq)
Na2CO3 (aq) + 2 HNO3 (aq)
SO3 (g) +
-----> 12 CO2 (g) + 11 H2O (R)
----->
MgCl2 (aq) + H2 (g)
----> 2 NaNO3 (aq) + H2CO3 (aq)
H2O (R) -----> H2SO4 (aq)
2 KHCO3 (s) ----> K2CO3 (s) +
CO2 (g) + H2O (g)
6.
a.How many moles of BaBr2 are contained in 150.0-g of BaBr2?
?molBaBr2 =
150gBaBr 1molBaBr
2 = 0.691molBaBr2
x
1
b.
217gBaBr2
How many atoms of Cl are in 42.6-g of CHCl3?
?atomsCl =
42.6gCHCl 3x6.022x1023 atomsCl
3x
= 6.44x1023 atomsCl
1
c.
119.5gCHCl3
How many grams of Sr are contained in sufficient SrCl2 to obtain 7.2 x
1022 atoms of Cl?
? gSr =
7.2x1022 atomsCl
1
d.
87.6gSr
2x6.022x1023 atomsCl
= 5.2gSr
How many grams of N are contained in 350 molecules of N2H4?
? gN =
e.
x
350moleculesN2 H4
1
x
28gN
23
6.022x10 moleculesN2 H4
= 1.6x10−20 gN
How many molecules of H2SO4 are contained in 75.0-g of H2SO4?
? moleculesH2SO4 =
75.0gH SO 6.022x1023 moleculesH SO
2
4x
2
4 = 4.61x1023 moleculesH2SO4
198gH2SO4
7.
a.How many grams of N are there in 15.0-g of NH4NO3?
? gN =
15.0gNH NO 28gN
4
3x
= 5.25gN
1
b.
80gNH4 NO3
How many molecules of water are in one gallon - (or 3.784 L). Assume the
density of water is 1 g/mL.
6.022x1023 moleculesHO
3.784L 1000mL 1g
? moleculesH2O =
= 1.266x1026 moleculesH2O
x
x
x
18gH2O
1
1L
1mL
2
c.
How many moles of C2H6 are in 115.0-g of C2H6?
? molC2 H6 =
115.0gC2 H6 1molC2 H6
= 3.833molC2 H6
x
1
d.
How many atoms of oxygen are contained in 245.0-g of KMnO4?
?atomsO =
245.0gKMnO 4x6.022x1023 atomsO
4x
= 3.736x1024 atomsO
1
e.
30gC2 H6
158gKMnO4
How many grams of BaCl2 are required to obtain 450 atoms of Cl?
=
? gramsBaCl
2
450atomsCl
1
x
208gramsBaCl2
23
2x6.022x10 atomsCl
= 7.8x10−20 gramsBaCl
2
8.
Hydrazine, N2H4, is used extensively as a rocket fuel. It can be prepared from
the reaction:
Given:
15.0-g NH 3
2 NH3 (g) +
Stoichiometry:
2 mol NH 3
2 x 17 g NH 3
17.5-g OClOCl - (aq) ----> N2H4 (aq)
1 mol OCl1 x 51.5 g OCl-
+
1 mol N 2 H 4
1 x 32 g N 2 H 4
Cl - (aq) +
1 mol Cl1 x 35.5 g Cl-
H2O (R)
1 mol H 2O
1 x 18 g H 2O
Suppose the reaction is carried out with 15.0-g of NH3 and 17.5-g of OCl -.
a.
Which is the limiting reactant?
15.0gNH3 = 0.441
17.5gOCl− = 0.340
1x51.5gOCl−
2x17gNH3
Since the stoichiometric ratio for OCl- is less than that for NH3, OCl- is the
limiting reactant.
Compare :
b.
How many grams hydrazine could be formed in the reaction above?
Based on limiting reactant:
? gN2 H4 =
17.5gOCl− 1x32gN H
2
4− = 10.9gN2 H4
x
1
c.
1x51.5gOCl
In the above reaction, how many grams of the excess reactant would be expected
to remain after the reaction is complete.
First, find out how many grams of the excess reactant, NH3, reacted.
−
? gNHreacted = 17.5gOCl x 2x17gNH3 = 11.6gNH
1x51.5gOCl−
1
3
3
The remaining amount is the starting amount minus the reacted amount:
15.0-g – 11.6g = 3.4g NH3 remain
9.
The reason hydrofluoric acid, HF(aq), cannot be stored in glass bottles and can be used in
etching glass is its reaction with silica, SiO2.
15.0-g SiO2
SiO2 (s)
1 mol SiO2
+
1 x 60 g SiO2
25.0-g HF
6 HF (aq) W
6 mol HF
6 x 20 g HF
H2SiF6 (aq) +
1 mol H2SiF
1 x 144 g H2SiF
2 H2O (l)
2 mol H2O
2 x 18 g H2O
Suppose 15.0-g of SiO2 are reacted with 25.0-g of HF.
a.
Which is the limiting reactant?
Using conversion factor approach:
15.0gSiO
? g HF required to use up SiO2 =
2
x
6x20gHF
= 30.0g HF required
11x60gSiO2
Since we only have 25.0-g of HF, we will run out and HF will be the limiting
reactant.
Alternatively, look at the two ratios formed from the equation above:
15.0gSiO2
25.0gHF
= 0.25
1x60gSiO
6x20gHF
= 0.208
2
Since the HF ratio is smaller, it is limiting using this approach, too, which is fortunate.
b.
How many grams of H2SiF6 could be produced in the reaction?
? g H2SiF6 formed =
25.0gHF 1x144g H SiF
2
6 = 30.0g H2SiF6 formed
x
16x20gHF
Notice this calculation is based on the limiting reactant, HF. It is also an
absolute coincidence that the grams of HF required and the grams of
H2SiF6 are the same
c.
If the experiment actually produces 22.5-g of H2SiF6 what is the percent yield?
% yield =
d.
actual yield
x100% = 22.5g x 100% = 75.0% yield
theoretical yield
30.0g
How many grams of the excess reactant remain at the end of the reaction?
? g SiO2used =
25.0gHF 1x60gSiO
2 = 12.5g SiO2 used
x
16x20gHF
#g SiO2 remaining at end = 15.0 g started with – 12.5 g used = 2.5 g SiO2
10.
The following reaction occurs during the production of ammonia.
55.0 g N2
20.0 g H2
3 H2 (g) W 2 NH3 (g)
N2 (g) +
1 mol N2
3 mol H2
2 mol NH3
1 x 28 g N2
3 x 2 g H2
2 x 17 g NH3
a.
Balance the equation.
b.
55.0-g of N2 are mixed with 20.0-g of H2. Which reactant is the limiting
reactant?
First, find out how much H2 is required to react with the N2, or vice versa if you wish. ?
g H2 required to react with 55.0g N2 =
55.0gN 3x2gH
2
x
1
2
= 11.8g H2 required
1x28gN2
Since we have 20.0 g of H 2 and only need 11.8 g, we have excess hydrogen and the
nitrogen limits the reaction.
c.
How many grams of ammonia could be produced from the reaction in part
b?
Using the limiting reactant, N2:
? g NH3 produced =
55.0gN 2x17gNH
2
1
d.
x
3
= 66.8g NH3 produced
1x28gN2
How many grams of the excess reactant would remain at the end of the
reaction?
From part b, we know that 11.8 g of H2 were used in the reaction. To
find the excess, simply subtract 11.8g from the starting amount.
#g H2 left over = 20.0 g – 11.8 g = 12.2 g H2
11.
Balance each of the following equations by placing coefficients in the appropriate places.
2 C6H6 (R) + 15O2
W 12 CO2 (g) +
4FeO (s) + 3 O2 (g) W
6 H2O (R)
2 Fe2O3 (s)
Au2S3(s) + 3 H2(g) W 2 Au (s) +3 H2S (g)
2 KClO3 (s) W
2KCl (s) +
2 Eu (s) + 6 HF (g) W
12.
3 O2 (g)
2 EuF3 (s) +
3 H2 (g)
Find the number of moles and molecules in 17.50-g of CO2. Be sure to show your work. ?
moles =
17.50gCO 1molCO
2x
2 = 0.3977molCO2
1
? molecules =
44gCO2
17.50gCO 6.022x1023 moleculesCO
2x
2 = 2.395x1023 molecules CO2
144gCO2
13.
Find the number of grams and moles in 4.5 x 1022 formula units of KBr.
14.
The reusable booster rockets of the space shuttle program use a mixture of aluminum
and ammonium perchlorate for fuel. A possible equation for this reaction is:
1000 g NH4ClO4
3Al (s) + 3NH4ClO4 (s) W
3 mol Al
3 mol NH4ClO4
3x27 g Al 3x117.5 g NH4ClO4
a.
b.
Al2O3 (s)
+ AlCl3 (s) + 3 NO (g) + 6 H2O (g)
3 mol NO 6 mol H2O
1 mol AlCl3
1 mol Al2O3
1x102 g Al2O3 1x133.5 g AlCl3 3x30 g NO 6 x 18 g H2O
What mass of aluminum is required to react with every 1000-g
of ammonium perchlorate burned?
1000gNH ClO 3x27gAl
4
4x
? gAl =
= 229.8gAl
1
3x117.5gNH4ClO4
What mass of NO is produced when the 1000-g of ammonium
perchlorate is burned?
3x30gNO
1000gNH4ClO4
x
? gNO =
= 255.3gNOproduced
3x117.5gNH4ClO4
1
15.
a.How many N atoms are contained in 75.0-g of N2H4 known as the rocket fuel
hydrazine?
?Natoms =
75.0gN H 2x6.022x1023 Natoms
2
4x
= 2.82x1024 Natoms
1
b.
32gN2 H4
How many hydrogen atoms are contained in 100 molecules of H2O2 ?
hydrogen atoms in each molecule, 100 molecules = 2 x 100 hydrogen atoms=
200 hydrogen atoms
c.
How many moles of CHF2Cl are contained in 150.0-g of CHF2Cl?
?molCHF2Cl =
150.0gCHF Cl 1molCHF Cl
2
2
x
= 1.734molCHF2Cl
186.5gCHF2Cl
16.
Ammonia can be converted to nitric oxide, NO, by the reaction below:
15.0 g NH3
a.
25.0 g O2
4 NH3 (g) +
5 O2 (g) ---->
4 NO (g)
4 mol NH3
5 mol O2
4 mol NO
4 x 17 g NH3
5 x 32 g O2
+ 6 H2O (g)
6 mol H2O
4 x 30 g NO 6 x 18 g H2O
15.0-g of NH3 are reacted with 25.0-g of O2. Which is the limiting
reactant?
Figure out how many grams of O2 are required to use up all of
the NH3.
? g O = 15.0 g NH3 x 5x32g O2 = 35.3g O required
2
2
4x17g NH3
1
Since we need 35.3 g of O2 to react with all of the NH3 and only
have 25.0g of O2, we will run out of O2. It is limiting.
b.
How many grams of NO could be formed in the reaction of part a?
?g NO =
25.0g O 4x30gNO
2x
= 18.8 g NO
1
c.
5x32g O2
How many grams of the excess reactant would remain after the reaction of
part a?
First, find out how many grams of the excess reactant, NH3, are used.
? g NH3 used =
25.0 g O2
x
4x17g NH3
= 10.6g NH3 used
5x32g O2
1
Amount of NH3 remaining = start amount of NH3 – amount of NH3 used
= 15.0 g – 10.6 g =4.4 g NH3
d.
If 15.0-g of NO is formed in the reaction of part a, what is the percent
yield?
% yield =
actual yield
x100% = 15.0gNO x100% = 79.8%
theoretical yield
18.8gNO
17.
Balance each of the following equations and classify it as combustion, decomposition,
combination, methathesis, and/or oxidation-reduction. Note that some equations may fall
into more than one category. State all categories that apply in each case.
Equation
2 C(gr) + ___ O2 (g) —> 2 CO (g)
Classification
combination,oxid-red
NH3 (g) + ___ HCl (g) —> ___ NH4Cl (s)
combination
Cr(s) + NiSO4 (aq) ----> Ni (s) + CrSO4 (aq)
oxidation-reduction
HC2H3O2 (aq) + NaOH(aq) —>HOH (R) + NaC2H3O2 (aq)
metathesis
2 H2O (R) —> 2H2 (g) + ___ O2 (g)
18.
decomposition,
oxidation-reduction
Answer each of the following questions. Show your work to receive full credit.
a. How many moles are contained in 14.5-g of CH3Br?
?mol CH3Br =
14.5g CH Br 1mol CH Br
3
3
x
= 0.153 mol CH3Br
1
95g CH3Br
b. How many oxygen atoms are contained in 5 molecules of C2H8O2?
? mol O atoms =
5 molecules C2 H8O2
2 O atoms
x
= 10 O atoms
1
1 molecule C H O
2
8 2
or just think about it – 2 O atoms per molecule and 5 molecules = 10 O atoms
c. How many grams are contained in 0.125-mol of Ba(NO3)2?
? g Ba(NO ) 2 = 0.125 moles Ba(NO3 )2 x 261 g Ba(NO3 )2 = 32.6 g Ba(NO ) 2
3
3
1 mole Ba(NO3 )2
1
d. What is the molar mass of MgSO4C7H2O?
molar mass = 1 x 24.3 + 1 x 32 + 4 x 16 + 7 x 18 = 246.3 g/mol
e. How many grams of carbon are contained in 25.0-g of CH2Cl2?
? g C = 25.0 g CH2Cl2 x 1x12 g C = 3.52g C
1
85gCH2Cl2
19.
Balance each of the following chemical equations by placing the appropriate
coefficients on the underlines.
___ TiCl4 + 4 H2O
___ P4
6 ___ TiO2
6
+ 10 Cl2
___ Al2O3 + 3 H2SeO4
2 C2H6 + 7 O2
6
6
4 PCl5
___ Al2(SeO4)3 + 3 H2O
4 CO2 + 6 H2O
___ Ca3(PO4)2 +2 H2SO4
20.
+ 4 HCl
6 ___ Ca(H2PO4)2
+ 2 CaSO4
Nitromethane, CH3NO2, reacts with oxygen to form carbon dioxide,
water, and nitrogen dioxide.
a. Write the balanced equation for this reaction.
4CH3NO2 + 7O2  4CO2 + 6H2O + 4NO2
b. How many moles of oxygen would be required to react with 8-mol of
nitromethane?
It takes 7 moles of O2 for every 4 mol of nitromethane, so 8 mol of nitromethane
would require 8 mol x 7/4 = 14 mol O2.
21.
The $-blocker drug, timolol, is expected to reduce the need for heart bypass
surgery. Its composition by mass is 47.2 %C, 6.55 %H, 13.0 %N, 25.9 %O, and
7.43 % S. Its molar mass is 432-g/mol.
a. What is the empirical formula for timolol?
Divide by smallest:
C: 47.2/12 = 3.93
3.93/0.232 = 16.9
H: 6.55/1 = 6.55
6.55/0.232 = 28.2
N: 13.0/14 = 0.929
0.929/0.232=4.00
O: 25.9/16 = 1.619
1.619/0.232=6.98
S: 7.43/32 = 0.232
0.232/0.232=1
C17H28N4O7S
Mass of this formula = 432
b. What is the molecular formula for timolol? Since the molar mass = mass of
the empirical formula, the empirical formula is the molecular formula:
C17H28N4O7S
22.
Find the percent mass of each element in C9H9N.
9x12
%C = 9x12 + 9x1+1x14 x100% = 82.4%C
9x1
%H = 9x12 + 9x1+1x14 x100% = 6.87%H
1x14
%N = 9x12 + 9x1+1x14 x100% = 10.7%N
23.
a. How many grams of NH3 are contained in 2.45 x 10-3 mol of NH3?
2.45x10−3 mol NH 17g NH
3x
3 = 0.0417g NH3
? g NH3 =
1
1 mol NH3
b. How many atoms of C are contained in 0.193-g of C12H22O11?
? atoms C =
0.193g C H O
12
22
11
x
12x6.022x1023 atoms C
= 4.08x1021 atoms C
1342g C12 H22O11
c. How many moles of PCl3 are contained in 75.0-g of PCl3?
? mol PCl3 =
75.0g PCl 1 mol PCl
3x
3 = 0.545 mol PCl3
1
137.5 g PCl3
d. How many atoms of oxygen are contained in 35 molecules of CO2?
35 molecules CO 2 atoms O
2x
?atoms O =
= 70 atoms O
1
1 molecule CO2
e. How many molecules of H2O are contained in 35.0-g of water?
35.0 g H O 6.022x1023 molecules H O
2
2
= 1.17x1024 molecules H2O
? molecules H2O =
x
118 g H2O
24.
a. How many grams of BaI2 are contained in 1.25-mol of BaI2?
? g BaI2 = 1.25 mol BaI2 x 391 g BaI2 = 489 g BaI2
1 mol BaI2
1
b. How many oxygen atoms are contained in 10.0-g of water?
?O atoms =
10.0 g H O 1x6.022x1023 O atoms
2
x
= 3.35x1023 O atoms
1
18 g H2O
c. How many moles of NH4NO3 are contained in 25.0-g of NH4NO3?
? mol NH4 NO3 =
25.0 g NH NO 1 mol NH NO
4
3x
4
3 = 0.312 mol NH4 NO3
180 g NH4 NO3