Solved Worksheet on Limiting Reactants Use the following | Chegg.com - Free Printable

Problem Analysis:

The worksheet involves two chemical reactions, and we need to solve for the moles of products formed and identify the limiting reactant in each case. Let's break it down step by step.

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Reaction 1: \( \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \)

#### a. How many moles of NH₃ can be produced from the reaction of 28 g of N₂?

1. Molar Mass Calculation:
- Molar mass of \( \text{N}_2 \): \( 2 \times 14.01 \, \text{g/mol} = 28.02 \, \text{g/mol} \).

2. Convert grams of N₂ to moles:
\[
\text{Moles of } \text{N}_2 = \frac{\text{Mass of } \text{N}_2}{\text{Molar mass of } \text{N}_2} = \frac{28 \, \text{g}}{28.02 \, \text{g/mol}} \approx 1.00 \, \text{mol}
\]

3. Stoichiometry:
- From the balanced equation: \( 1 \, \text{mol of } \text{N}_2 \) produces \( 2 \, \text{mol of } \text{NH}_3 \).
- Therefore, \( 1.00 \, \text{mol of } \text{N}_2 \) will produce:
\[
1.00 \, \text{mol of } \text{N}_2 \times \frac{2 \, \text{mol of } \text{NH}_3}{1 \, \text{mol of } \text{N}_2} = 2.00 \, \text{mol of } \text{NH}_3
\]

Answer: \( \boxed{2.00} \) moles of \( \text{NH}_3 \).

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#### b. How many moles of NH₃ can be produced from the reaction of 25 g of H₂?

1. Molar Mass Calculation:
- Molar mass of \( \text{H}_2 \): \( 2 \times 1.01 \, \text{g/mol} = 2.02 \, \text{g/mol} \).

2. Convert grams of H₂ to moles:
\[
\text{Moles of } \text{H}_2 = \frac{\text{Mass of } \text{H}_2}{\text{Molar mass of } \text{H}_2} = \frac{25 \, \text{g}}{2.02 \, \text{g/mol}} \approx 12.38 \, \text{mol}
\]

3. Stoichiometry:
- From the balanced equation: \( 3 \, \text{mol of } \text{H}_2 \) produces \( 2 \, \text{mol of } \text{NH}_3 \).
- Therefore, \( 12.38 \, \text{mol of } \text{H}_2 \) will produce:
\[
12.38 \, \text{mol of } \text{H}_2 \times \frac{2 \, \text{mol of } \text{NH}_3}{3 \, \text{mol of } \text{H}_2} \approx 8.25 \, \text{mol of } \text{NH}_3
\]

Answer: \( \boxed{8.25} \) moles of \( \text{NH}_3 \).

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#### c. If 28 g of N₂ and 25 g of H₂ are reacted together, which one would be the limiting reactant?

1. From part (a):
- \( 28 \, \text{g of } \text{N}_2 \) produces \( 2.00 \, \text{mol of } \text{NH}_3 \).

2. From part (b):
- \( 25 \, \text{g of } \text{H}_2 \) produces \( 8.25 \, \text{mol of } \text{NH}_3 \).

3. Determine the limiting reactant:
- The reactant that produces the smaller amount of product is the limiting reactant.
- Here, \( \text{N}_2 \) produces only \( 2.00 \, \text{mol of } \text{NH}_3 \), while \( \text{H}_2 \) could theoretically produce \( 8.25 \, \text{mol of } \text{NH}_3 \).
- Therefore, \( \text{N}_2 \) is the limiting reactant.

Answer: \( \boxed{\text{N}_2} \).

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Reaction 2: \( \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{H}_2\text{O} + 4\text{CO}_2 \)

#### a. How many moles of water can be produced from the reaction of 28 g of C₃H₈?

1. Molar Mass Calculation:
- Molar mass of \( \text{C}_3\text{H}_8 \): \( (3 \times 12.01) + (8 \times 1.01) = 36.03 + 8.08 = 44.11 \, \text{g/mol} \).

2. Convert grams of C₃H₈ to moles:
\[
\text{Moles of } \text{C}_3\text{H}_8 = \frac{\text{Mass of } \text{C}_3\text{H}_8}{\text{Molar mass of } \text{C}_3\text{H}_8} = \frac{28 \, \text{g}}{44.11 \, \text{g/mol}} \approx 0.635 \, \text{mol}
\]

3. Stoichiometry:
- From the balanced equation: \( 1 \, \text{mol of } \text{C}_3\text{H}_8 \) produces \( 3 \, \text{mol of } \text{H}_2\text{O} \).
- Therefore, \( 0.635 \, \text{mol of } \text{C}_3\text{H}_8 \) will produce:
\[
0.635 \, \text{mol of } \text{C}_3\text{H}_8 \times \frac{3 \, \text{mol of } \text{H}_2\text{O}}{1 \, \text{mol of } \text{C}_3\text{H}_8} \approx 1.905 \, \text{mol of } \text{H}_2\text{O}
\]

Answer: \( \boxed{1.91} \) moles of \( \text{H}_2\text{O} \).

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#### b. How many moles of water can be produced from the reaction of 45 g of O₂?

1. Molar Mass Calculation:
- Molar mass of \( \text{O}_2 \): \( 2 \times 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol} \).

2. Convert grams of O₂ to moles:
\[
\text{Moles of } \text{O}_2 = \frac{\text{Mass of } \text{O}_2}{\text{Molar mass of } \text{O}_2} = \frac{45 \, \text{g}}{32.00 \, \text{g/mol}} \approx 1.406 \, \text{mol}
\]

3. Stoichiometry:
- From the balanced equation: \( 5 \, \text{mol of } \text{O}_2 \) produces \( 3 \, \text{mol of } \text{H}_2\text{O} \).
- Therefore, \( 1.406 \, \text{mol of } \text{O}_2 \) will produce:
\[
1.406 \, \text{mol of } \text{O}_2 \times \frac{3 \, \text{mol of } \text{H}_2\text{O}}{5 \, \text{mol of } \text{O}_2} \approx 0.844 \, \text{mol of } \text{H}_2\text{O}
\]

Answer: \( \boxed{0.844} \) moles of \( \text{H}_2\text{O} \).

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#### c. If 28 g of C₃H₈ and 45 g of O₂ are reacted together, which one would be the limiting reactant?

1. From part (a):
- \( 28 \, \text{g of } \text{C}_3\text{H}_8 \) produces \( 1.91 \, \text{mol of } \text{H}_2\text{O} \).

2. From part (b):
- \( 45 \, \text{g of } \text{O}_2 \) produces \( 0.844 \, \text{mol of } \text{H}_2\text{O} \).

3. Determine the limiting reactant:
- The reactant that produces the smaller amount of product is the limiting reactant.
- Here, \( \text{O}_2 \) produces only \( 0.844 \, \text{mol of } \text{H}_2\text{O} \), while \( \text{C}_3\text{H}_8 \) could theoretically produce \( 1.91 \, \text{mol of } \text{H}_2\text{O} \).
- Therefore, \( \text{O}_2 \) is the limiting reactant.

Answer: \( \boxed{\text{O}_2} \).

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Final Answers:

1. a. \( \boxed{2.00} \)
b. \( \boxed{8.25} \)
c. \( \boxed{\text{N}_2} \)

2. a. \( \boxed{1.91} \)
b. \( \boxed{0.844} \)
c. \( \boxed{\text{O}_2} \)