Here are the step-by-step solutions for each problem on the worksheet.
1. Reaction: Aluminum and Oxygen
First, we must balance the chemical equation to find the mole ratios.
Balanced Equation: $4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3$
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Ratio of Al to $\text{Al}_2\text{O}_3$: 4 moles of Al produce 2 moles of $\text{Al}_2\text{O}_3$.
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Ratio of Al to $\text{O}_2$: 4 moles of Al react with 3 moles of $\text{O}_2$.
a. How many moles of aluminum oxide will be formed from 17 moles of aluminum?
$$17 \text{ mol Al} \times \frac{2 \text{ mol Al}_2\text{O}_3}{4 \text{ mol Al}} = 8.5 \text{ mol Al}_2\text{O}_3$$
b. How many moles of oxygen are needed to react with 23.8 moles of aluminum?
$$23.8 \text{ mol Al} \times \frac{3 \text{ mol O}_2}{4 \text{ mol Al}} = 17.85 \text{ mol O}_2$$
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2. Reaction: Ammonia and Oxygen
Equation: $4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O}$ (Already balanced)
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Ratio of $\text{NH}_3$ to $\text{O}_2$: 4 : 5
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Ratio of $\text{O}_2$ to $\text{H}_2\text{O}$: 5 : 6
a. How many moles of oxygen are needed to react with 3.24 moles of ammonia ($\text{NH}_3$)?
$$3.24 \text{ mol NH}_3 \times \frac{5 \text{ mol O}_2}{4 \text{ mol NH}_3} = 4.05 \text{ mol O}_2$$
b. How many moles of water are produced from 12.8 moles of oxygen?
$$12.8 \text{ mol O}_2 \times \frac{6 \text{ mol H}_2\text{O}}{5 \text{ mol O}_2} = 15.36 \text{ mol H}_2\text{O}$$
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3. Reaction: Iron(III) Oxide and Carbon Monoxide
Equation: $\text{Fe}_3\text{O}_4 + 4\text{CO} \rightarrow 3\text{Fe} + 4\text{CO}_2$ (Already balanced)
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Ratio of $\text{Fe}_3\text{O}_4$ to $\text{CO}_2$: 1 : 4
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Ratio of $\text{Fe}_3\text{O}_4$ to CO: 1 : 4
a. How many moles of carbon dioxide are produced from 2.87 moles of $\text{Fe}_3\text{O}_4$?
$$2.87 \text{ mol Fe}_3\text{O}_4 \times \frac{4 \text{ mol CO}_2}{1 \text{ mol Fe}_3\text{O}_4} = 11.48 \text{ mol CO}_2$$
b. How many moles of carbon monoxide are needed to react with 8.25 moles of $\text{Fe}_3\text{O}_4$?
$$8.25 \text{ mol Fe}_3\text{O}_4 \times \frac{4 \text{ mol CO}}{1 \text{ mol Fe}_3\text{O}_4} = 33.0 \text{ mol CO}$$
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4. Reaction: Lead(II) Oxide and Oxygen
Equation: $6\text{PbO} + \text{O}_2 \rightarrow 2\text{Pb}_3\text{O}_4$ (Already balanced)
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Ratio of $\text{O}_2$ to $\text{Pb}_3\text{O}_4$: 1 : 2
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Ratio of PbO to $\text{O}_2$: 6 : 1
a. How many moles of $\text{Pb}_3\text{O}_4$ are produced from 1.25 moles of oxygen?
$$1.25 \text{ mol O}_2 \times \frac{2 \text{ mol Pb}_3\text{O}_4}{1 \text{ mol O}_2} = 2.50 \text{ mol Pb}_3\text{O}_4$$
b. How many moles of oxygen must react with 8.75 moles of lead(II) oxide?
$$8.75 \text{ mol PbO} \times \frac{1 \text{ mol O}_2}{6 \text{ mol PbO}} \approx 1.46 \text{ mol O}_2$$
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Final Answer:
1.
a. 8.5 moles
b. 17.85 moles
2.
a. 4.05 moles
b. 15.36 moles
3.
a. 11.48 moles
b. 33.0 moles
4.
a. 2.50 moles
b. 1.46 moles
Parent Tip: Review the logic above to help your child master the concept of mole to mole stoichiometry worksheet.
