Worksheet titled "Gas Stoichiometry Worksheet" featuring six problems related to gas laws and stoichiometry, including calculations of moles, mass, volume, and reactions at STP.

Gas Stoichiometry Worksheet Solutions

#### Molar Volume Problems

1. Calculate the number of moles contained in 550 mL of carbon dioxide at STP. (Answer: 0.0246 mol)

- Given:
- Volume of CO₂ = 550 mL = 0.550 L (since 1 L = 1000 mL)
- STP conditions: 1 mole of any gas occupies 22.4 L.

- Formula:
\[
\text{Number of moles} = \frac{\text{Volume of gas at STP}}{\text{Molar volume at STP}}
\]
\[
\text{Number of moles} = \frac{0.550 \, \text{L}}{22.4 \, \text{L/mol}}
\]

- Calculation:
\[
\text{Number of moles} = \frac{0.550}{22.4} \approx 0.0246 \, \text{mol}
\]

- Answer: \( \boxed{0.0246 \, \text{mol}} \)

2. Calculate the mass of 1.50 L of CH₄ at STP. (Answer: 1.07 g)

- Given:
- Volume of CH₄ = 1.50 L
- Molar mass of CH₄ = 12.01 g/mol (C) + 4 × 1.01 g/mol (H) = 16.05 g/mol
- STP conditions: 1 mole of any gas occupies 22.4 L.

- Steps:
1. Calculate the number of moles of CH₄:
\[
\text{Number of moles} = \frac{\text{Volume of gas at STP}}{\text{Molar volume at STP}}
\]
\[
\text{Number of moles} = \frac{1.50 \, \text{L}}{22.4 \, \text{L/mol}}
\]
\[
\text{Number of moles} = \frac{1.50}{22.4} \approx 0.06696 \, \text{mol}
\]

2. Calculate the mass of CH₄:
\[
\text{Mass} = \text{Number of moles} \times \text{Molar mass}
\]
\[
\text{Mass} = 0.06696 \, \text{mol} \times 16.05 \, \text{g/mol}
\]
\[
\text{Mass} \approx 1.07 \, \text{g}
\]

- Answer: \( \boxed{1.07 \, \text{g}} \)

3. Calculate the volume in liters of 50.0 grams of nitrogen dioxide at STP. (Answer: 24.3 L)

- Given:
- Mass of NO₂ = 50.0 g
- Molar mass of NO₂ = 14.01 g/mol (N) + 2 × 16.00 g/mol (O) = 46.01 g/mol
- STP conditions: 1 mole of any gas occupies 22.4 L.

- Steps:
1. Calculate the number of moles of NO₂:
\[
\text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}}
\]
\[
\text{Number of moles} = \frac{50.0 \, \text{g}}{46.01 \, \text{g/mol}}
\]
\[
\text{Number of moles} \approx 1.087 \, \text{mol}
\]

2. Calculate the volume of NO₂ at STP:
\[
\text{Volume} = \text{Number of moles} \times \text{Molar volume at STP}
\]
\[
\text{Volume} = 1.087 \, \text{mol} \times 22.4 \, \text{L/mol}
\]
\[
\text{Volume} \approx 24.3 \, \text{L}
\]

- Answer: \( \boxed{24.3 \, \text{L}} \)

#### Gas Stoichiometry Problems

4. Given the following reaction: NH₄NO₂(s) → N₂(g) + 2H₂O(g). How many liters of nitrogen gas is produced if 50.0 L of water is produced at STP? (Answer: 25.0 L)

- Given:
- Volume of H₂O produced = 50.0 L
- Balanced equation: NH₄NO₂(s) → N₂(g) + 2H₂O(g)
- STP conditions: 1 mole of any gas occupies 22.4 L.

- Steps:
1. From the balanced equation, the stoichiometric ratio is:
\[
1 \, \text{mole of N₂} : 2 \, \text{moles of H₂O}
\]
This means that for every 2 moles of H₂O produced, 1 mole of N₂ is produced.

2. Calculate the number of moles of H₂O produced:
\[
\text{Number of moles of H₂O} = \frac{\text{Volume of H₂O at STP}}{\text{Molar volume at STP}}
\]
\[
\text{Number of moles of H₂O} = \frac{50.0 \, \text{L}}{22.4 \, \text{L/mol}}
\]
\[
\text{Number of moles of H₂O} \approx 2.232 \, \text{mol}
\]

3. Using the stoichiometric ratio, calculate the number of moles of N₂ produced:
\[
\text{Number of moles of N₂} = \frac{1}{2} \times \text{Number of moles of H₂O}
\]
\[
\text{Number of moles of N₂} = \frac{1}{2} \times 2.232 \approx 1.116 \, \text{mol}
\]

4. Calculate the volume of N₂ produced at STP:
\[
\text{Volume of N₂} = \text{Number of moles of N₂} \times \text{Molar volume at STP}
\]
\[
\text{Volume of N₂} = 1.116 \, \text{mol} \times 22.4 \, \text{L/mol}
\]
\[
\text{Volume of N₂} \approx 25.0 \, \text{L}
\]

- Answer: \( \boxed{25.0 \, \text{L}} \)

5. Given the following reaction: CuO(s) + H₂(g) → Cu(s) + H₂O(g). If 250. L of hydrogen gas are used to reduce copper(II) oxide at STP, what mass of copper is to be expected? (Answer: 709 g)

- Given:
- Volume of H₂ used = 250. L
- Balanced equation: CuO(s) + H₂(g) → Cu(s) + H₂O(g)
- Molar mass of Cu = 63.55 g/mol
- STP conditions: 1 mole of any gas occupies 22.4 L.

- Steps:
1. Calculate the number of moles of H₂ used:
\[
\text{Number of moles of H₂} = \frac{\text{Volume of H₂ at STP}}{\text{Molar volume at STP}}
\]
\[
\text{Number of moles of H₂} = \frac{250. \, \text{L}}{22.4 \, \text{L/mol}}
\]
\[
\text{Number of moles of H₂} \approx 11.16 \, \text{mol}
\]

2. From the balanced equation, the stoichiometric ratio is:
\[
1 \, \text{mole of H₂} : 1 \, \text{mole of Cu}
\]
This means that for every 1 mole of H₂ used, 1 mole of Cu is produced.

3. Calculate the number of moles of Cu produced:
\[
\text{Number of moles of Cu} = \text{Number of moles of H₂}
\]
\[
\text{Number of moles of Cu} = 11.16 \, \text{mol}
\]

4. Calculate the mass of Cu produced:
\[
\text{Mass of Cu} = \text{Number of moles of Cu} \times \text{Molar mass of Cu}
\]
\[
\text{Mass of Cu} = 11.16 \, \text{mol} \times 63.55 \, \text{g/mol}
\]
\[
\text{Mass of Cu} \approx 709 \, \text{g}
\]

- Answer: \( \boxed{709 \, \text{g}} \)

6. How many grams of solid lithium must be added to liquid water in order to obtain 15.0 L of hydrogen gas at STP? (Answer: 9.29 g)

- Given:
- Volume of H₂ produced = 15.0 L
- Balanced equation: 2 Li + 2H₂O → 2 LiOH + H₂
- Molar mass of Li = 6.94 g/mol
- STP conditions: 1 mole of any gas occupies 22.4 L.

- Steps:
1. Calculate the number of moles of H₂ produced:
\[
\text{Number of moles of H₂} = \frac{\text{Volume of H₂ at STP}}{\text{Molar volume at STP}}
\]
\[
\text{Number of moles of H₂} = \frac{15.0 \, \text{L}}{22.4 \, \text{L/mol}}
\]
\[
\text{Number of moles of H₂} \approx 0.6696 \, \text{mol}
\]

2. From the balanced equation, the stoichiometric ratio is:
\[
2 \, \text{moles of Li} : 1 \, \text{mole of H₂}
\]
This means that for every 1 mole of H₂ produced, 2 moles of Li are required.

3. Calculate the number of moles of Li required:
\[
\text{Number of moles of Li} = 2 \times \text{Number of moles of H₂}
\]
\[
\text{Number of moles of Li} = 2 \times 0.6696 \approx 1.3392 \, \text{mol}
\]

4. Calculate the mass of Li required:
\[
\text{Mass of Li} = \text{Number of moles of Li} \times \text{Molar mass of Li}
\]
\[
\text{Mass of Li} = 1.3392 \, \text{mol} \times 6.94 \, \text{g/mol}
\]
\[
\text{Mass of Li} \approx 9.29 \, \text{g}
\]

- Answer: \( \boxed{9.29 \, \text{g}} \)

Final Answers:

1. \( \boxed{0.0246 \, \text{mol}} \)
2. \( \boxed{1.07 \, \text{g}} \)
3. \( \boxed{24.3 \, \text{L}} \)
4. \( \boxed{25.0 \, \text{L}} \)
5. \( \boxed{709 \, \text{g}} \)
6. \( \boxed{9.29 \, \text{g}} \)