Chemistry – Stoichiometry Worksheet
1.
How many grams of calcium carbonate are required to prepare 50.0 g of calcium
oxide?
___CaCO3  ___CaO + ___CO2
50.0 / 56.077 = 0.892 mol CaO = 0.892 mol CaCO3
0.892 (100.086) = 89.3 g of CaCO3
2.
Using the reaction from #1 determine how many moles of carbon dioxide will be
produced from the decomposition of 150.00 g of calcium carbonate.
150.00 / 100.086 = 1.4987 mol CaCO3 = 1.4987 mol CO2
When the coefficients are the same the moles will be the same
3.
When 0.500 g of magnesium reacts with silver nitrate, how many grams of silver
are prepared?
___Mg + _2_AgNO3  _2_Ag + ___Mg(NO3)2
0.500 / 24.305 = 0.0206 mol Mg
2Ag
0.0206
= 0.0412 mol Ag
1Mg
0.0412 (107.87) = 4.44 g of Ag
4.
Based on the reaction in #3, calculate the mass of silver nitrate needed to
generate 100.00 g of silver.
100.00 / 107.87 = 0.92704 mol of Ag = 0.92704 mol AgNO3
0.92704 (169.87) = 157.48 g of AgNO3
5.
If 75.0 g of copper react with mercuric nitrate, how many grams of mercury
form?
Cu + Hg(NO3)2  Cu(NO3)2 + Hg
75.0 / 63.546 = 1.18 mol Cu = 1.18 mol Hg
1.18 (200.59) = 237 g of Hg
6.
When 60.0 g of aluminum react with hydrochloric acid, how many grams of
hydrochloric acid react?
_2_Al + _6_HCl  _2_AlCl3 + _3_H2
60.0 / 26.982 = 2.22 mol Al
6HCl
2.22
= 6.66 mol HCl
2Al
6.66 (36.461) = 243 g of HCl
7.
Use the reaction from problem #6 to calculate the mass of each reactant needed
to produce 12.54 moles of hydrogen gas.
2Al
= 8.360 mol Al
3H 2
8.360 (26.982) = 225.6 g Al
6HCl
12.54
= 25.08 mol HCl
3H 2
25.08 (36.461) = 914 g HCl
12.54
8.
How many grams of magnesium chloride are produced by treating 4.00 g of
titanium (III) chloride with magnesium?
_2_TiCl3 + _3_Mg  _3_ MgCl2 + _2_Ti
4.00 / 154.226 = 0.0259 mol TiCl3
3MgCl2
0.0259
= 0.0389 mol MgCl2
2TiCl3
0.0389 (95.211) = 3.70 g MgCl2
9.
What mass of Na2SO4 is produced when sulfuric acid reacts with 200.0 g of
sodium chloride?
_2_NaCl + ___H2SO4  _2_HCl + ___Na2SO4
200.0 / 58.443 = 3.422 mol NaCl
1Na2 SO4
3.422
= 1.711 mol Na2SO4
2NaCl
1.711 (142.041) = 243.0 g of Na2SO4
10. From the reaction in #9 calculate how many moles of sulfuric acid, H2SO4, is
required to produce 125.00 g of hydrochloric acid, HCl.
125.00 / 36.461 = 3.4283 mol HCl
1H 2 SO4
3.4283
= 1.7142 mol H2SO4
2HCl
1.7142 (98.077) = 168.12 g of H2SO4