Q: In the reaction $$2\mathrm{Fe} + 3\mathrm{S} \rightarrow \mathrm{Fe}_2\mathrm{S}_3$$, how many moles of sulfur are required to react completely with 1.42 moles of iron?

$$1.42 \text{ mol} \times \frac{3 \text{ mol S}}{2 \text{ mol Fe}} = 2.13 \text{ mol S}$$

Q: Propane ($$\mathrm{C}_3\mathrm{H}_8) $$combusts according to the balanced equation: $$\mathrm{C}_3\mathrm{H}_8 + 5\mathrm{O}_2 \rightarrow 3\mathrm{CO}_2 + 4\mathrm{H}_2\mathrm{O}. $$How many moles of $$\mathrm{CO}_2$$ are produced from 2.25 mol of $$\mathrm{C}_3\mathrm{H}_8$$?

$$2.25 \text{ mol} \times \frac{3 \text{ mol CO}_2}{1 \text{ mol C}_3\mathrm{H}_8} = 6.75 \text{ mol CO}_2$$

Q: Given the unbalanced reaction $$\mathrm{N}_2 + \mathrm{O}_2 \rightarrow \mathrm{N}_2\mathrm{O}_4$$, if 15.0 g of $$\mathrm{N}_2\mathrm{O}_4 is $$produced, how many moles of $$\mathrm{O}_2$$ were required? (Molar mass of $$\mathrm{N}_2\mathrm{O}_4 = 92.0 g/$$mol)

$$15.0 \text{ g} \div 92.0 \text{ g/mol} = 0.163 \text{ mol N}_2\mathrm{O}_4$$; 0.163 mol $$\mathrm{O}_2$$ required

Q: When 54.6 g of acetylene ($$\mathrm{C}_2\mathrm{H}_2) is $$burned in excess oxygen, how many grams of $$\mathrm{CO}_2$$ are produced? (Molar mass $$\mathrm{C}_2\mathrm{H}_2 = 26.0 g/$$mol, $$\mathrm{CO}_2 = 44.0 g/$$mol; reaction: $$2\mathrm{C}_2\mathrm{H}_2 + 5\mathrm{O}_2 \rightarrow 4\mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O})$$

$$54.6 \text{ g} \div 26.0 \text{ g/mol} = 2.10 \text{ mol C}_2\mathrm{H}_2$$; $$2.10 \text{ mol} \times \frac{4}{2} = 4.20 \text{ mol CO}_2$$; $$4.20 \text{ mol} \times 44.0 \text{ g/mol} = 184.8 \text{ g CO}_2$$

Q: Heptane ($$\mathrm{C}_7\mathrm{H}_{16}) $$combusts according to $$\mathrm{C}_7\mathrm{H}_{16} + 11\mathrm{O}_2 \rightarrow 7\mathrm{CO}_2 + 8\mathrm{H}_2\mathrm{O}. $$How many grams of $$\mathrm{O}_2$$ are required to react with 22.50 g of heptane? (Molar mass $$\mathrm{C}_7\mathrm{H}_{16} = 100.0 g/$$mol, $$\mathrm{O}_2 = 32.0 g/$$mol)

$$22.50 \text{ g} \div 100.0 \text{ g/mol} = 0.225 \text{ mol C}_7\mathrm{H}_{16}$$; $$0.225 \text{ mol} \times 11 = 2.475 \text{ mol O}_2$$; $$2.475 \text{ mol} \times 32.0 \text{ g/mol} = 79.2 \text{ g O}_2$$

Q: If 31.1 g of Fe3O4 is produced in the reaction above, and the theoretical yield is 34.6 g, what is the percent yield?

$$\frac{31.1}{34.6} \times 100\% = 89.9\%$$

Q: How many moles of $$\mathrm{Cl}_2$$ are needed to react with 4.40 mol of Al according to the balanced equation $$2\mathrm{Al} + 3\mathrm{Cl}_2 \rightarrow 2\mathrm{AlCl}_3$$?

$$4.40 \text{ mol Al} \times \frac{3 \text{ mol Cl}_2}{2 \text{ mol Al}} = 6.6 \text{ mol Cl}_2$$

Question 1

Which of the following best defines the term 'limiting reactant' in a chemical reaction?

Accepted Answer

The reactant that is completely consumed first, thus limiting the amount of product formed.

Question 2

In the reaction $$2\mathrm{Fe} + 3\mathrm{S} \rightarrow \mathrm{Fe}_2\mathrm{S}_3$$, how many moles of sulfur are required to react completely with 1.42 moles of iron?

Accepted Answer

$$1.42 	ext{ mol} 	imes \frac{3 	ext{ mol S}}{2 	ext{ mol Fe}} = 2.13 	ext{ mol S}$$

Question 3

Propane ($$\mathrm{C}_3\mathrm{H}_8) $$combusts according to the balanced equation: $$\mathrm{C}_3\mathrm{H}_8 + 5\mathrm{O}_2 \rightarrow 3\mathrm{CO}_2 + 4\mathrm{H}_2\mathrm{O}. $$How many moles of $$\mathrm{CO}_2$$ are produced from 2.25 mol of $$\mathrm{C}_3\mathrm{H}_8$$?

Accepted Answer

$$2.25 	ext{ mol} 	imes \frac{3 	ext{ mol CO}_2}{1 	ext{ mol C}_3\mathrm{H}_8} = 6.75 	ext{ mol CO}_2$$

Question 4

Given the unbalanced reaction $$\mathrm{N}_2 + \mathrm{O}_2 \rightarrow \mathrm{N}_2\mathrm{O}_4$$, if 15.0 g of $$\mathrm{N}_2\mathrm{O}_4 is $$produced, how many moles of $$\mathrm{O}_2$$ were required? (Molar mass of $$\mathrm{N}_2\mathrm{O}_4 = 92.0 g/$$mol)

Accepted Answer

$$15.0 	ext{ g} \div 92.0 	ext{ g/mol} = 0.163 	ext{ mol N}_2\mathrm{O}_4$$; 0.163 mol $$\mathrm{O}_2$$ required

Question 5

When 54.6 g of acetylene ($$\mathrm{C}_2\mathrm{H}_2) is $$burned in excess oxygen, how many grams of $$\mathrm{CO}_2$$ are produced? (Molar mass $$\mathrm{C}_2\mathrm{H}_2 = 26.0 g/$$mol, $$\mathrm{CO}_2 = 44.0 g/$$mol; reaction: $$2\mathrm{C}_2\mathrm{H}_2 + 5\mathrm{O}_2 \rightarrow 4\mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O})$$

Accepted Answer

$$54.6 	ext{ g} \div 26.0 	ext{ g/mol} = 2.10 	ext{ mol C}_2\mathrm{H}_2$$; $$2.10 	ext{ mol} 	imes \frac{4}{2} = 4.20 	ext{ mol CO}_2$$; $$4.20 	ext{ mol} 	imes 44.0 	ext{ g/mol} = 184.8 	ext{ g CO}_2$$

Question 6

Heptane ($$\mathrm{C}_7\mathrm{H}_{16}) $$combusts according to $$\mathrm{C}_7\mathrm{H}_{16} + 11\mathrm{O}_2 \rightarrow 7\mathrm{CO}_2 + 8\mathrm{H}_2\mathrm{O}. $$How many grams of $$\mathrm{O}_2$$ are required to react with 22.50 g of heptane? (Molar mass $$\mathrm{C}_7\mathrm{H}_{16} = 100.0 g/$$mol, $$\mathrm{O}_2 = 32.0 g/$$mol)

Accepted Answer

$$22.50 	ext{ g} \div 100.0 	ext{ g/mol} = 0.225 	ext{ mol C}_7\mathrm{H}_{16}$$; $$0.225 	ext{ mol} 	imes 11 = 2.475 	ext{ mol O}_2$$; $$2.475 	ext{ mol} 	imes 32.0 	ext{ g/mol} = 79.2 	ext{ g O}_2$$

Question 7

Which of the following best describes 'percent yield' in a chemical reaction?

Accepted Answer

The ratio of the actual yield to the theoretical yield, multiplied by 100%.

Question 8

In a reaction, 25.0 g of Fe and 40.0 g of H2O are reacted to form Fe3O4 and H2. If the balanced equation is $$3\mathrm{Fe} + 4\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Fe}_3\mathrm{O}_4 + 4\mathrm{H}_2$$, which reactant is limiting? (Molar masses: Fe = 55.8 g/mol, H2O = 18.0 g/mol)

Accepted Answer

H2O is the limiting reactant.

Question 9

If 31.1 g of Fe3O4 is produced in the reaction above, and the theoretical yield is 34.6 g, what is the percent yield?

Accepted Answer

$$\frac{31.1}{34.6} 	imes 100\% = 89.9\%$$

Question 10

Aluminum reacts with chlorine gas to form aluminum chloride. What is the balanced chemical equation for this reaction?

Accepted Answer

$$2\mathrm{Al} + 3\mathrm{Cl}_2 \rightarrow 2\mathrm{AlCl}_3$$

Question 11

How many moles of $$\mathrm{Cl}_2$$ are needed to react with 4.40 mol of Al according to the balanced equation $$2\mathrm{Al} + 3\mathrm{Cl}_2 \rightarrow 2\mathrm{AlCl}_3$$?

Accepted Answer

$$4.40 	ext{ mol Al} 	imes \frac{3 	ext{ mol Cl}_2}{2 	ext{ mol Al}} = 6.6 	ext{ mol Cl}_2$$

Question 12

In a reaction, 45.0 g of Al and 62.0 g of Cl2 are mixed to form AlCl3. If 66.5 g of AlCl3 is actually obtained, what is the percent yield? (Molar masses: Al = 27.0 g/mol, Cl2 = 71.0 g/mol, AlCl3 = 133.5 g/mol)

Accepted Answer

First, find limiting reactant: $$45.0 	ext{ g Al} \div 27.0 	ext{ g/mol} = 1.67 	ext{ mol Al}$$; $$62.0 	ext{ g Cl}_2 \div 71.0 	ext{ g/mol} = 0.873 	ext{ mol Cl}_2$$. $$2\mathrm{Al} + 3\mathrm{Cl}_2 ightarrow 2\mathrm{AlCl}_3$$. $$1.67 	ext{ mol Al} 	imes \frac{3}{2} = 2.505 	ext{ mol Cl}_2$$ needed, but only 0.873 mol available, so Cl2 is limiting. $$0.873 	ext{ mol Cl}_2 	imes \frac{2 	ext{ mol AlCl}_3}{3 	ext{ mol Cl}_2} = 0.582 	ext{ mol AlCl}_3$$; $$0.582 	ext{ mol} 	imes 133.5 	ext{ g/mol} = 77.7 	ext{ g}$$ theoretical yield. Percent yield: $$\frac{66.5}{77.7} 	imes 100\% = 85.6\%$$

Question 13

A student mixes 10.0 g of $$\mathrm{CaCl}_2$$ with 10.0 g of $$\mathrm{Na}_2\mathrm{CO}_3 in $$water, producing $$\mathrm{CaCO}_3$$ and $$\mathrm{NaCl}. If $$the theoretical yield of $$\mathrm{CaCO}_3 is 12.5 g $$and the actual yield is 11.0 g, what is the percent yield?

Accepted Answer

$$\frac{11.0}{12.5} 	imes 100\% = 88.0\%$$

Question 14

Which of the following statements about the excess reactant is correct?

Accepted Answer

It is the reactant that remains after the limiting reactant is used up.

Question 15

A chemical reaction has a theoretical yield of 185 g and an actual yield of 79.8 g. What is the percent yield?

Accepted Answer

$$\frac{79.8}{185} 	imes 100\% = 43.1\%$$

Question 16

In the reaction $$2\mathrm{NO}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HNO}_3 + \mathrm{NO}$$, how many grams of $$\mathrm{HNO}_3$$ are produced from 8.25 g of $$\mathrm{NO}_2$$? (Molar mass $$\mathrm{NO}_2 = 46.0 g/$$mol, $$\mathrm{HNO}_3 = 63.0 g/$$mol)

Accepted Answer

$$8.25 	ext{ g} \div 46.0 	ext{ g/mol} = 0.179 	ext{ mol NO}_2$$; $$0.179 	ext{ mol} 	imes \frac{1 	ext{ mol HNO}_3}{2 	ext{ mol NO}_2} = 0.0895 	ext{ mol HNO}_3$$; $$0.0895 	ext{ mol} 	imes 63.0 	ext{ g/mol} = 5.64 	ext{ g HNO}_3$$

Stoichiometry and Limiting Reactant Study Guide – Step-by-Step Guidance

스터디 가이드 - 연습 문제

학습 가이드 - 플래시카드