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)
C3H8O2 (R) + 4 O2 (g) --------> 3 CO2 (g) +
+ 3 BaSO4 (s)
4 H2O (g)
P2O5 (g) + 3 H2O (R) ----> 2 H3PO4 (aq)
B10H18 + 12 O2 -----> 5B2O3
+
9 H2O
N2(g) + 3 H2 (g) -----> 2 NH3 (g)
2.
a.
How many grams of Cl are in 113.5-g of C2Cl6?
? gCl =
b.
113.5 gC2Cl6
213gCl
= 102.0 gCl
x
1
237 gC2Cl6
How many atoms of O are in 45.0-g of C12H22O11?
45.0 gC12 H 22O11 11x6.022 x1023 atomsO
? atomsO =
x
= 8.72 x1023 atomsO
1
342 gC12 H 22O11
c.
How many moles of CaBr2 are required to obtain 4.5 x 1021 atoms of Br?
? molCaBr2 =
d.
4.5 x1021 atomsBr
1molCaBr2
x
= 3.7 x10−3 molCaBr2
23
1
2 x6.022 x10 atomsBr
How many grams of C are required to make 1.35-mol of CH4?
? gC =
e.
1.35molCH 4
12 gC
x
= 16.2 gC
1
1molCH 4
How many moles of NH3 are contained in 250.0-g of NH3?
? molNH 3 =
250.0 gNH 3 1molNH 3
x
= 14.7molNH 3
1
17 gNH 3
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
2 x 267 g AlBr3
6 mol KBr
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?
Compare
20.0 gK 2CO3
= 0.0483
3 x138 gK 2CO3
30.0 gAlBr3
= 0.0562
2 x 267 gAlBr3
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?
20.0 gK 2CO3 1x 234 gAl2 (CO3 )3
x
= 11.3gAl2 (CO3 )
1
3x138 gK 2CO3
Note this calculation is based on the limiting reactant, not the excess.
? gAl2 (CO3 )3 =
d.
Suppose the reaction is carried out and 7.5-g of Al2(CO3)3 are formed. What is the
percent yield?
% yield =
e.
7.5 gAl2 (CO3 )3 Produced
x100% = 66.4% yield
11.3gAl2 (CO3 )3 Calculated
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 =
20.0 gK 2CO3 2 x 267 gAlBr3
x
= 25.8 gAlBr3used
1
3x138 gK 2CO3
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:
7.1
= 7.1
1
7.1
= 12.91
0.55
O:
26.2
= 1.6375
16
1.6375
= 2.98
0.55
N:
7.7
= 0.55
14
0.55
=1
0.55
4.92
= 8.95
0.55
Divide by smallest (0.55) :
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)
Mg (s)
+
+ 12 O2 -----> 12 CO2 (g) + 11 H2O (R)
2HCl (aq) ----->
Na2CO3 (aq) + 2 HNO3 (aq)
SO3 (g) +
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 =
b.
How many atoms of Cl are in 42.6-g of CHCl3?
? atomsCl =
c.
42.6 gCHCl3 3 x6.022 x10 23 atomsCl
x
= 6.44 x1023 atomsCl
1
119.5 gCHCl3
How many grams of Sr are contained in sufficient SrCl2 to obtain 7.2 x 1022
atoms of Cl?
? gSr =
d.
7.2 x10 22 atomsCl
87.6 gSr
x
= 5.2 gSr
1
2 x6.022 x1023 atomsCl
How many grams of N are contained in 350 molecules of N2H4?
? gN =
e.
150 gBaBr 1molBaBr2
x
= 0.691molBaBr2
1
217 gBaBr2
350moleculesN 2 H 4
28 gN
x
= 1.6 x10−20 gN
23
1
6.022 x10 moleculesN 2 H 4
How many molecules of H2SO4 are contained in 75.0-g of H2SO4?
? moleculesH 2 SO4 =
75.0 gH 2 SO4 6.022 x10 23 moleculesH 2 SO4
x
= 4.61x10 23 moleculesH 2 SO4
1
98 gH 2 SO4
7.
a.
How many grams of N are there in 15.0-g of NH4NO3?
? gN =
b.
How many molecules of water are in one gallon - (or 3.784 L). Assume the
density of water is 1 g/mL.
? moleculesH 2O =
c.
15.0 gNH 4 NO3
28 gN
x
= 5.25 gN
1
80 gNH 4 NO3
3.784 L 1000mL 1g 6.022 x10 23 moleculesH 2O
x
x
x
= 1.266 x1026 moleculesH 2O
1
1L
1mL
18 gH 2O
How many moles of C2H6 are in 115.0-g of C2H6?
? molC2 H 6 =
d.
115.0 gC2 H 6 1molC2 H 6
x
= 3.833molC2 H 6
1
30 gC2 H 6
How many atoms of oxygen are contained in 245.0-g of KMnO4?
245.0 gKMnO4 4 x6.022 x1023 atomsO
? atomsO =
x
= 3.736 x10 24 atomsO
1
158 gKMnO4
e.
How many grams of BaCl2 are required to obtain 450 atoms of Cl?
? gramsBaCl2 =
450atomsCl
208 gramsBaCl2
x
= 7.8 x10−20 gramsBaCl2
23
1
2 x6.022 x10 atomsCl
8.
Given:
Hydrazine, N2H4, is used extensively as a rocket fuel. It can be prepared from the
reaction:
15.0-g NH3 17.5-g OCl-
2 NH3 (g)
+
Stoichiometry:
2 mol NH3
2 x 17 g NH3
OCl - (aq) ----> N2H4 (aq)
1 mol OCl1 x 51.5 g OCl-
1 mol N2H4
1 x 32 g N2H4
+
Cl - (aq)
+
H2O (R)
1 mol Cl1 mol H2O
1 x 35.5 g Cl- 1 x 18 g H2O
Suppose the reaction is carried out with 15.0-g of NH3 and 17.5-g of OCl -.
a.
Which is the limiting reactant?
15.0 gNH 3
17.5 gOCl −
= 0.441
= 0.340
2 x17 gNH 3
1x51.5 gOCl −
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:
? gN 2 H 4 =
c.
17.5 gOCl − 1x32 gN 2 H 4
x
= 10.9 gN 2 H 4
1
1x51.5 gOCl −
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.
? gNH 3reacted =
17.5 gOCl −
2 x17 gNH 3
x
= 11.6 gNH 3
1
1x51.5 gOCl −
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:
15.0 gSiO2 6 x 20 gHF
? g HF required to use up SiO 2 =
x
= 30.0 g HF required
1
1x60 gSiO2
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:
15.0 gSiO2
25.0 gHF
= 0.25
= 0.208
1x60 gSiO2
6 x 20 gHF
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?
25.0 gHF 1x144 g H 2 SiF6
x
= 30.0 g H 2 SiF6 formed
1
6 x 20 gHF
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
H 2 SiF6 are the same
? g H 2 SiF6 formed =
c.
If the experiment actually produces 22.5-g of H2SiF6 what is the percent yield?
% yield =
d.
actual yield
22.5 g
x100% =
x 100% = 75.0% yield
theoretical yield
30.0 g
How many grams of the excess reactant remain at the end of the reaction?
? g SiO2used =
25.0 gHF 1x60 gSiO2
x
= 12.5 g SiO2 used
1
6 x 20 gHF
#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
N2 (g) +
1 mol N2
1 x 28 g N2
20.0 g H2
3 H2 (g)
3 mol H2
3 x 2 g H2
W
2 NH3 (g)
2 mol NH3
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.
55.0 gN 2 3x 2 gH 2
? g H 2 required to react with 55.0 g N 2 =
x
= 11.8 g H 2 required
1
1x 28 gN 2
Since we have 20.0 g of H2 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 NH 3 produced =
d.
55.0 gN 2 2 x17 gNH 3
x
= 66.8 g NH 3 produced
1
1x 28 gN 2
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) + 15 O2 W
12 CO2 (g) +
4 FeO (s) + 3 O2 (g)
2 KCl (s) +
2 Eu (s) + 6 HF (g) W
12.
W 2 Fe2O3 (s)
W 2 Au (s) + 3 H2S (g)
Au2S3(s) + 3 H2(g)
2 KClO3 (s) W
6 H2O (R)
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 =
? molecules =
17.50 gCO2 1molCO2
x
= 0.3977molCO2
1
44 gCO2
17.50 gCO2 6.022 x10 23 moleculesCO2
x
= 2.395 x10 23 molecules CO2
1
44 gCO2
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)
1 mol Al2O3
1 mol AlCl3
3 mol NO 6 mol H2O
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?
1000 gNH 4ClO 4
3x 27 gAl
? gAl =
x
= 229.8 gAl
1
3x117.5 gNH 4ClO 4
What mass of NO is produced when the 1000-g of ammonium perchlorate
is burned?
1000 gNH 4ClO 4
3x30 gNO
? gNO =
x
= 255.3gNOproduced
1
3x117.5 gNH 4ClO 4
15.
a.
How many N atoms are contained in 75.0-g of N2H4 known as the rocket fuel
hydrazine?
? Natoms =
b.
75.0 gN 2 H 4 2 x6.022 x10 23 Natoms
x
= 2.82 x1024 Natoms
1
32 gN 2 H 4
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 =
150.0 gCHF2Cl 1molCHF2Cl
x
= 1.734molCHF2Cl
1
86.5 gCHF2Cl
16.
Ammonia can be converted to nitric oxide, NO, by the reaction below:
25.0 g O2
15.0 g NH3
4 NH3 (g) + 5 O2 (g) ----> 4 NO (g) + 6 H2O (g)
4 mol NH3
4 x 17 g NH3
a.
5 mol O2
5 x 32 g O2
4 mol NO
4 x 30 g NO
6 mol H2O
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 O2 =
15.0 g NH 3 5 x32 g O2
x
= 35.3g O2 required
1
4 x17 g NH 3
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 =
c.
25.0 g O2 4 x30 gNO
x
= 18.8 g NO
1
5 x32 g 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 NH 3 used =
25.0 g O2 4 x17 g NH 3
x
= 10.6 g NH 3 used
1
5 x32 g O2
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
15.0 gNO
x100% =
x100% = 79.8%
theoretical yield
18.8 gNO
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)
NH3 (g) + ___ HCl (g) —> ___ NH4Cl (s)
Cr(s) + NiSO4 (aq) ----> Ni (s) + CrSO4 (aq)
HC2H3O2 (aq) + NaOH(aq) —>HOH (R) + NaC2H3O2 (aq)
2 H2O (R) —> 2H2 (g) + ___ O2 (g)
18.
Classification
combination,oxid-red
combination
oxidation-reduction
metathesis
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 CH 3 Br =
14.5 g CH 3 Br 1mol CH 3 Br
x
= 0.153 mol CH 3 Br
1
95 g CH 3 Br
b. How many oxygen atoms are contained in 5 molecules of C2H8O2?
? mol O atoms =
5 molecules C2 H 8O2
2 O atoms
x
= 10 O atoms
1
1 molecule C2 H 8O2
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( NO3 ) 2 =
0.125 moles Ba ( NO3 ) 2 261 g Ba( NO3 ) 2
x
= 32.6 g Ba( NO3 ) 2
1
1 mole Ba( NO3 ) 2
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 CH 2Cl 2 1x12 g C
x
= 3.52 g C
1
85 gCH 2Cl2
19.
Balance each of the following chemical equations by placing the appropriate
coefficients on the underlines.
___ TiCl4 + 4 H2O
___ P4
6
6
+ 10 Cl2
___ Al2O3 + 3 H2SeO4
2 C2H6 + 7 O2
6
___ TiO2
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.
4 CH3NO2 + 7 O2 Æ 4 CO2 + 6 H2O + 4 NO2
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.
9 x12
x100% = 82.4%C
9 x12 + 9 x1 + 1x14
9 x1
%H =
x100% = 6.87% H
9 x12 + 9 x1 + 1x14
1x14
%N =
x100% = 10.7% N
9 x12 + 9 x1 + 1x14
%C =
23.
a. How many grams of NH3 are contained in 2.45 x 10-3 mol of NH3?
2.45 x10 −3 mol NH 3 17 g NH 3
? g NH 3 =
x
= 0.0417 g NH 3
1
1 mol NH 3
b. How many atoms of C are contained in 0.193-g of C12H22O11?
0.193g C12 H 22O11 12 x6.022 x10 23 atoms C
? atoms C =
x
= 4.08 x10 21 atoms C
1
342 g C12 H 22O11
c. How many moles of PCl3 are contained in 75.0-g of PCl3?
75.0 g PCl3 1 mol PCl3
? mol PCl3 =
x
= 0.545 mol PCl3
1
137.5 g PCl3
d. How many atoms of oxygen are contained in 35 molecules of CO2?
35 molecules CO2
2 atoms O
? atoms O =
x
= 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 2O 6.022 x1023 molecules H 2O
? molecules H 2O =
x
= 1.17 x1024 molecules H 2O
1
18 g H 2O
24.
a. How many grams of BaI2 are contained in 1.25-mol of BaI2?
1.25 mol BaI 2 391 g BaI 2
? g BaI 2 =
x
= 489 g BaI 2
1
1 mol BaI 2
b. How many oxygen atoms are contained in 10.0-g of water?
10.0 g H 2O 1x6.022 x1023 O atoms
? O atoms =
x
= 3.35 x1023 O atoms
1
18 g H 2O
c. How many moles of NH4NO3 are contained in 25.0-g of NH4NO3?
25.0 g NH 4 NO3 1 mol NH 4 NO3
? mol NH 4 NO3 =
x
= 0.312 mol NH 4 NO3
1
80 g NH 4 NO3