To solve the problem, we will follow these steps:
### Step 1: Write the balanced chemical equation
The reaction between aluminum and sulfuric acid can be represented as:
\[
2 \text{Al} + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3 \text{H}_2
\]
### Step 2: Calculate moles of aluminum
Given the mass of aluminum (Al) is 5.4 g and the molar mass of aluminum is 27.0 g/mol, we can calculate the moles of aluminum:
\[
\text{Moles of Al} = \frac{\text{mass}}{\text{molar mass}} = \frac{5.4 \, \text{g}}{27.0 \, \text{g/mol}} = 0.2 \, \text{mol}
\]
### Step 3: Calculate moles of sulfuric acid
We have 50.0 mL of 5.0 M sulfuric acid (H₂SO₄). First, convert the volume from mL to L:
\[
50.0 \, \text{mL} = 0.050 \, \text{L}
\]
Now, calculate the moles of sulfuric acid:
\[
\text{Moles of H}_2\text{SO}_4 = \text{concentration} \times \text{volume} = 5.0 \, \text{mol/L} \times 0.050 \, \text{L} = 0.25 \, \text{mol}
\]
### Step 4: Determine the limiting reagent
From the balanced equation, we see that:
- 2 moles of Al react with 3 moles of H₂SO₄.
Now, calculate the required moles of H₂SO₄ for 0.2 moles of Al:
\[
\text{Required moles of H}_2\text{SO}_4 = 0.2 \, \text{mol Al} \times \frac{3 \, \text{mol H}_2\text{SO}_4}{2 \, \text{mol Al}} = 0.3 \, \text{mol H}_2\text{SO}_4
\]
We have 0.25 moles of H₂SO₄ available, which is less than the 0.3 moles required. Therefore, H₂SO₄ is the limiting reagent.
### Step 5: Calculate moles of hydrogen produced
According to the balanced equation, 3 moles of H₂SO₄ produce 3 moles of H₂. Thus, the moles of hydrogen produced from 0.25 moles of H₂SO₄ is:
\[
\text{Moles of H}_2 = 0.25 \, \text{mol H}_2\text{SO}_4 \times \frac{3 \, \text{mol H}_2}{3 \, \text{mol H}_2\text{SO}_4} = 0.25 \, \text{mol H}_2
\]
### Step 6: Calculate the volume of hydrogen gas produced
Using the ideal gas law \( PV = nRT \), we can rearrange to find the volume \( V \):
\[
V = \frac{nRT}{P}
\]
Where:
- \( n = 0.25 \, \text{mol} \)
- \( R = 0.082 \, \text{atm L mol}^{-1} K^{-1} \)
- \( T = 300 \, K \)
- \( P = 1.0 \, \text{atm} \)
Substituting the values:
\[
V = \frac{0.25 \, \text{mol} \times 0.082 \, \text{atm L mol}^{-1} K^{-1} \times 300 \, K}{1.0 \, \text{atm}}
\]
\[
V = \frac{6.15 \, \text{L}}{1.0} = 6.15 \, \text{L}
\]
### Final Answer:
The volume of hydrogen gas produced is **6.15 L**.
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