Gas stoichiometry practice worksheet with five problems on calculating volumes and masses in chemical reactions.

Problem: Gas Stoichiometry Practice 1.0

The task involves solving gas stoichiometry problems using molar volume relationships. Let's solve each problem step by step.

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#### 1. Calculate the volume of oxygen gas required to react with 2.5 L of hydrogen gas at STP, assuming no change in temperature or pressure.

Given:
- Volume of hydrogen gas (\( H_2 \)) = 2.5 L at STP.
- Balanced chemical equation:
\[
2H_2(g) + O_2(g) \rightarrow 2H_2O(g)
\]

Step 1: Understand the relationship between volumes at STP.
At STP (Standard Temperature and Pressure), one mole of any gas occupies 22.4 L. The balanced equation shows that 2 moles of \( H_2 \) react with 1 mole of \( O_2 \). Therefore, the volume ratio of \( H_2 \) to \( O_2 \) is also 2:1.

Step 2: Use the volume ratio to find the volume of \( O_2 \).
From the equation:
\[
\text{Volume ratio of } H_2 : O_2 = 2 : 1
\]
If 2.5 L of \( H_2 \) is given, the volume of \( O_2 \) required is:
\[
\text{Volume of } O_2 = \frac{\text{Volume of } H_2}{2} = \frac{2.5 \, \text{L}}{2} = 1.25 \, \text{L}
\]

Answer:
\[
\boxed{1.25 \, \text{L}}
\]

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#### 2. What volume of \( CO_2 \) gas will be produced from 17.5 L of octane (\( C_8H_{18} \)) reacting with excess oxygen?

Given:
- Volume of octane (\( C_8H_{18} \)) = 17.5 L at STP.
- Balanced chemical equation for the combustion of octane:
\[
C_8H_{18}(g) + \frac{25}{2}O_2(g) \rightarrow 8CO_2(g) + 9H_2O(g)
\]

Step 1: Understand the relationship between volumes at STP.
At STP, the volume of a gas is directly proportional to the number of moles. From the balanced equation, 1 mole of \( C_8H_{18} \) produces 8 moles of \( CO_2 \). Therefore, the volume ratio of \( C_8H_{18} \) to \( CO_2 \) is 1:8.

Step 2: Use the volume ratio to find the volume of \( CO_2 \).
If 17.5 L of \( C_8H_{18} \) is given, the volume of \( CO_2 \) produced is:
\[
\text{Volume of } CO_2 = 8 \times \text{Volume of } C_8H_{18} = 8 \times 17.5 \, \text{L} = 140 \, \text{L}
\]

Answer:
\[
\boxed{140 \, \text{L}}
\]

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#### 3. What volume of hydrogen gas at STP can be produced from the reaction of 36.5 g of zinc with hydrochloric acid?

Given:
- Mass of zinc (\( Zn \)) = 36.5 g.
- Balanced chemical equation:
\[
Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)
\]

Step 1: Calculate the moles of zinc.
The molar mass of zinc (\( Zn \)) is 65.38 g/mol. The number of moles of zinc is:
\[
\text{Moles of } Zn = \frac{\text{Mass of } Zn}{\text{Molar mass of } Zn} = \frac{36.5 \, \text{g}}{65.38 \, \text{g/mol}} \approx 0.558 \, \text{mol}
\]

Step 2: Use the stoichiometry of the reaction.
From the balanced equation, 1 mole of \( Zn \) produces 1 mole of \( H_2 \). Therefore, 0.558 mol of \( Zn \) will produce 0.558 mol of \( H_2 \).

Step 3: Convert moles of \( H_2 \) to volume at STP.
At STP, 1 mole of any gas occupies 22.4 L. The volume of \( H_2 \) produced is:
\[
\text{Volume of } H_2 = \text{Moles of } H_2 \times 22.4 \, \text{L/mol} = 0.558 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 12.5 \, \text{L}
\]

Answer:
\[
\boxed{12.5 \, \text{L}}
\]

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#### 4. How many grams of \( NaCl \) can be produced by the reaction of 1.56 g of sodium at STP with an excess of chlorine?

Given:
- Mass of sodium (\( Na \)) = 1.56 g.
- Balanced chemical equation:
\[
2Na(s) + Cl_2(g) \rightarrow 2NaCl(s)
\]

Step 1: Calculate the moles of sodium.
The molar mass of sodium (\( Na \)) is 22.99 g/mol. The number of moles of sodium is:
\[
\text{Moles of } Na = \frac{\text{Mass of } Na}{\text{Molar mass of } Na} = \frac{1.56 \, \text{g}}{22.99 \, \text{g/mol}} \approx 0.0675 \, \text{mol}
\]

Step 2: Use the stoichiometry of the reaction.
From the balanced equation, 2 moles of \( Na \) produce 2 moles of \( NaCl \). Therefore, 0.0675 mol of \( Na \) will produce 0.0675 mol of \( NaCl \).

Step 3: Convert moles of \( NaCl \) to grams.
The molar mass of \( NaCl \) is 58.44 g/mol. The mass of \( NaCl \) produced is:
\[
\text{Mass of } NaCl = \text{Moles of } NaCl \times \text{Molar mass of } NaCl = 0.0675 \, \text{mol} \times 58.44 \, \text{g/mol} \approx 3.95 \, \text{g}
\]

Answer:
\[
\boxed{3.95 \, \text{g}}
\]

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#### 5. In order to produce 10.0 liters of ammonia gas (\( NH_3 \)) at STP, how many grams of both nitrogen and hydrogen would you need to react?

Given:
- Volume of \( NH_3 \) = 10.0 L at STP.
- Balanced chemical equation:
\[
N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)
\]

Step 1: Calculate the moles of \( NH_3 \).
At STP, 1 mole of any gas occupies 22.4 L. The number of moles of \( NH_3 \) is:
\[
\text{Moles of } NH_3 = \frac{\text{Volume of } NH_3}{22.4 \, \text{L/mol}} = \frac{10.0 \, \text{L}}{22.4 \, \text{L/mol}} \approx 0.446 \, \text{mol}
\]

Step 2: Use the stoichiometry of the reaction.
From the balanced equation, 1 mole of \( N_2 \) and 3 moles of \( H_2 \) produce 2 moles of \( NH_3 \). Therefore:
- Moles of \( N_2 \) required:
\[
\text{Moles of } N_2 = \frac{\text{Moles of } NH_3}{2} = \frac{0.446 \, \text{mol}}{2} \approx 0.223 \, \text{mol}
\]
- Moles of \( H_2 \) required:
\[
\text{Moles of } H_2 = \frac{3 \times \text{Moles of } NH_3}{2} = \frac{3 \times 0.446 \, \text{mol}}{2} \approx 0.669 \, \text{mol}
\]

Step 3: Convert moles to grams.
- Molar mass of \( N_2 \) = 28.02 g/mol:
\[
\text{Mass of } N_2 = \text{Moles of } N_2 \times \text{Molar mass of } N_2 = 0.223 \, \text{mol} \times 28.02 \, \text{g/mol} \approx 6.25 \, \text{g}
\]
- Molar mass of \( H_2 \) = 2.02 g/mol:
\[
\text{Mass of } H_2 = \text{Moles of } H_2 \times \text{Molar mass of } H_2 = 0.669 \, \text{mol} \times 2.02 \, \text{g/mol} \approx 1.35 \, \text{g}
\]

Answer:
\[
\boxed{6.25 \, \text{g of } N_2 \text{ and } 1.35 \, \text{g of } H_2}
\]

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Final Answers:

1. \(\boxed{1.25 \, \text{L}}\)
2. \(\boxed{140 \, \text{L}}\)
3. \(\boxed{12.5 \, \text{L}}\)
4. \(\boxed{3.95 \, \text{g}}\)
5. \(\boxed{6.25 \, \text{g of } N_2 \text{ and } 1.35 \, \text{g of } H_2}\)