Stoichiometry Worksheet: Name: _________ANSWERS________________________
1. If you had 5.6 g of CaO & unlimited H2O, how many grams of Ca(OH)2 could you get?
CaO + H2O  Ca(OH )2
5.6 g/ 56 g/m of CaO = 0.1 m of CaO
(1 mol Ca(OH)2 / 1 mol CaO)(0.1 m CaO) = 0.1 m Ca(OH)2
0.1 m Ca(OH)2 x 74 g/m = 7.4 grams of Ca(OH)2
2. How many moles of LiOH is needed to make 5 moles of Fe(OH)3?
Fe(NO3)3 + 3 LiOH  3 LiNO3 + Fe(OH)3
[3 m LiOH/ 1 m Fe(OH)3] [ 5 m Fe(OH)3] = 15 mol LiOH
3. One kilogram of Cu with 22.4 moles of nitric acid can make how many moles of NO?
3 Cu + 8 HNO3  3 Cu(NO3)2 + 4 H2O + 2 NO
(Must balance equation 1st)
[2 mol NO/ 8 mol HNO3] [ 22.4 mol HNO3] = 5.6 mol NO
[3 mol Cu/ 8 mol HNO3] [ 22.4 mol HNO3] = 8.4 mol Cu
1000 g Cu / 63.5 g/m Cu = 15.7 mol Cu
[2 mol NO/ 3 mol Cu] [ 15.7 mol Cu] = 10.5 mol NO
4. You can produce how many grams of water by burning 54 L of acetylene (C2H2) in the presence of
166 L of oxygen?
2 C2 H2 + 5 O2  4 CO2 + 2 H2O
[22.4 L/mole for gases]
166 L O2/ 22.4 L/mol = 7.4 mol O2
54L C2H2 / 22.4 L/mol = 2.4 mol C2H2
[5 mol O2/ 2 mol C2H2] [ 2.4 mol C2H2] = 6 mol O2
[2 mol H2O/ 2 mol C2H2] [2.4 mol C2H2] = 2.4 mol H2O
(2.4 mol H2O)(18 g/mol) = 43.2 g H2O
5. Given: Al2O3 + 3 H2SO4  Al2(SO4)3 + 3 H2O
If I needed to produce 2.2 kg of aluminum
sulfate, how many grams of aluminum oxide would I need to react with an unlimited amount
of sulfuric acid?
2200 g / 342 g/mol = 6.4 mol Al2(SO4)3
(6.4 mol Al2O3)(102 g/mol) = 652.8 g Al2O3
6. Given: C12 H22 O11 + 12 O2(g)  12 CO2(g) + 11 H2O; Gases’ standard molar volume is 22.4L/mol.
If I combust 1 kg of the above carbohydrate, how many liters of CO2 will I produce?
(12g/m)(12) + (1g/m)(22) + (16 g/m)(11) = 342 g/m for C12 H22 O11
1000 g/ 342 g/m = 2.9 mol C12 H22 O11
(12 mol CO2/ 1 mol C12 H22 O11)(2.9 mol C12 H22 O11) = 35 mol CO2
(22.4 L/mol)(35 mol CO2) = 786 L CO2
7. If I used 5.6 g of copper in 50 mL of a 0.10 M solution of AgNO3, how much silver could I
precipitate, given the following equation? Cu + 2 AgNO3  Cu(NO3)2 + 2 Ag
(5.6 g/ 63.5 g/mol) = 0.088 mol Cu
(0.050L)(0.10 mol/L) = 0.005 mol AgNO3
(1 mol Cu/ 2 mol AgNO3) (0.005 mol AgNO3) = 0.0025 mol Cu
(2 mol Ag/ 2 mol AgNO3) (0.005 mol AgNO3) = 0.005 mol Ag
(0.005 mol Ag)(108 g/mol) = 0.54 g Ag
8. Given: Fe2S+ 3 O2 (g)  Fe2O3 + SO3
produced from 0.5 kg of Fe2S?
How many grams of iron(III) oxide could be
Fe2S: (56 g/mol)(2) + 32 g/m = 144 g/mol
Fe2O3: (56 g/mol)(2) + (16 g/mol)(3) = 160 g/mol
500 g/ 144 g/mol = 3.47 mol Fe2S
(1 mol Fe2O3/ 1 mol Fe2S)(3.47 mol Fe2S) = 3.47 mol Fe2O3
(3.47 mol)(160 g/mol) = 555 g Fe2O3