1. Write the balanced chemical equation

Aluminium reacts with sulphuric acid as:

$2\text{Al} + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3\text{H}_2$

2. Calculate moles of aluminium

Given mass of Al = $5.4\,\text{g}$

Molar mass of Al = $27.0\,\text{g mol}^{-1}$

$n(\text{Al}) = \frac{5.4}{27.0} = 0.20\,\text{mol}$

3. Calculate moles of sulphuric acid

Volume of $\text{H}_2\text{SO}_4$ = $50.0\,\text{mL} = 0.0500\,\text{L}$

Molarity = $5.0\,\text{M}$

$n(\text{H}_2\text{SO}_4) = M \times V = 5.0 \times 0.0500 = 0.25\,\text{mol}$

4. Find the limiting reagent

From the balanced equation:

$2\text{ mol Al} : 3\text{ mol H}_2\text{SO}_4$

For $0.20$ mol Al, required $\text{H}_2\text{SO}_4$ is:

$0.20 \times \frac{3}{2} = 0.30\,\text{mol}$

Available $\text{H}_2\text{SO}_4 = 0.25\,\text{mol}$

Since available acid is less than required, $\text{H}_2\text{SO}_4$ is the limiting reagent.

5. Calculate moles of hydrogen produced

From the equation:

$3\text{H}_2\text{SO}_4 \rightarrow 3\text{H}_2$

So,

$n(\text{H}_2) = n(\text{H}_2\text{SO}_4) = 0.25\,\text{mol}$

6. Use ideal gas equation to find volume

$PV = nRT$

Given:

• $P = 1.0\,\text{atm}$
• $n = 0.25\,\text{mol}$
• $R = 0.082\,\text{atm L mol}^{-1}\text{K}^{-1}$
• $T = 300\,\text{K}$

$V = \frac{nRT}{P} = \frac{0.25 \times 0.082 \times 300}{1.0}$

$V = 6.15\,\text{L}$

7. Final answer

The volume of hydrogen gas produced is:

$\boxed{6.15\,\text{L}}$