Exercise 12.2a
Stoichiometry I - Answers
DIRECTIONS: On a separate sheet of paper, solve the following problems.
1.
Given the equation NH3 + HCl NH4Cl, find the number of formula units of NH4Cl formed if 65.0 g of NH3 reacts with
an excess of HCl.
1NH3 + 1HCl → 1NH4Cl
Analyze, rewrite, balance equation
65.0 g
? p.
Solve for only reactant
2.
65.0 g NH3
x
6.02 x1023 p. NH4Cl
1 mol NH4Cl
= 2.30 x 1024 p. NH4Cl
x
1 mol WO3
6.02 x1023 p. WO3
x
1 mol W
1 mol WO3
183.84g W
1 mol W
x
= 12 300 g W
7.02 x 1014 p. CrCl3
1 mol CrCl3
6.02 x1023 p. CrCl3
x
2 mol CrCl2
2 mol CrCl3
x
x
6.02 x1023 p. CrCl2
1 mol CrCl2
= 7.02 x 1014 p. CrCl2
1 mol RbCl
6.02 x1023 p. RbCl
x
1 mol RbClO4
1 mol RbCl
x
x
184.917g RbClO4
1 mol RbClO4
= 0.154 g RbClO4
Given the equation C3H5(NO3)3 N2 + O2 + CO2 + H2O, find the mass of CO2 formed if 5.62 g C3H5(NO3)3 decomposes.
4C3H5(NO3)3 → 6N2 + 1O2 + 12CO2 + 10H2O
Analyze, rewrite, balance equation
5.62 g
?g
5.62 g C3H5(NO3)3
x
1 mol C3H5(NO3)3
227.085 g C3H5(NO3)3
x
12 mol CO2
4 mol C3H5(NO3)3
x
44.009g CO2
1 mol CO2
= 3.27 g CO2
Given the equation NH3 + HC1  NH4Cl, find the mass of NH3 needed to form 15.0 g of NH4Cl.
1NH3 + 1HCl → 1NH4Cl
Analyze, rewrite, balance equation
?g
15g
Solve for only reactant
7.
4.03 x 1025 p. WO3
5.00 x 1020 p. RbCl
Solve for only reactant
6.
1 mol NH4Cl
1 mol NH3
Given the equation RbCl + O2  RbClO4, find the mass of RbClO4 formed if 5.00 x 1020 formula units of RbCl reacts
with an excess of O2.
1RbCl
+ 2O2 → 1RbClO4
Analyze, rewrite, balance equation
5.00 x 1020 p.
?g
Solve for only reactant
5.
x
Given the equation Zn + CrCl3 CrCl2 + ZnCl2, find the number of formula units of CrCl2 formed if 7.02 x 1014 formula
units of CrCl3 reacts with an excess of Zn.
1Zn +
2CrCl3
→ 2CrCl2 + 1ZnCl2
Analyze, rewrite, balance equation
7.02 x 1014 p.
? p.
Solve for only reactant
4.
1 mol NH3
17.031g NH3
Given the equation WO3 + H2  W + H2O, find the mass of W formed if 4.03 x 10 25 formula units of WO3 reacts with an
excess of H2.
1WO3
+ 3H2 → 1W + 3H2O
Analyze, rewrite, balance equation
4.03 x 1025 p.
?g
Solve for only reactant
3.
x
15.0 g NH4Cl
x
1 mol NH4Cl
53.492g NH4Cl
x
1 mol NH3
1 mol NH4Cl
x
17.031g NH3
1 mol NH3
= 4.78 g NH3
Given the equation WO3 + H2  W + H2O, find the mass of WO3 needed to form 7.30 x 1021 formula units of W.
1WO3 + 3H2 →
1W
+ 3H2O
Analyze, rewrite, balance equation
?g
7.3 x 1021 p.
Solve for only reactant
7.30 x 1021 p. W
x
1 mol W
6.02 x1023 p. W
x
1 mol WO3
1 mol W
x
231.837g WO3
1 mol WO3
= 2.81 g WO3
Exercise 12.2a
Stoichiometry I - Answers
8.
Given the equation Zn + CrCl3  CrCl2 + ZnCl2, find the mass of ZnCl2 formed if 1.20 x 1024 formula units of CrCl3
reacts with an excess of Zn.
1Zn +
2CrCl3
→ 2CrCl2 + 1ZnCl2
Analyze, rewrite, balance equation
1.2 x 1024 p.
.
?g
Solve for only reactant
9.
1.20 x 1024 p. CrCl3
1 mol CrCl3
6.02 x1023 p. CrCl3
x
x
1 mol ZnCl2
2 mol CrCl3
x
136.286g ZnCl2
1 mol ZnCl2
= 136 g ZnCl2
Given the equation RbCl + O2  RbClO4, find the mass of O2 necessary to completely react 87.5 g RbCl.
1RbCl + 2O2 → 1RbClO4
Analyze, rewrite, balance equation
87.5 g
?g
Solve for only reactant
87.5 g RbCl
x
1 mol RbCl
120.921 g RbCl
x
2 mol O2
1 mol RbCl
x
31.998g O2
1 mol O2
= 46.3 g O2