Question

Aluminium reacts with sulfuric acid to form aluminium sulfate and hydrogen. What is the volume of hydrogen gas in liters $( L )$ produced at $300 \,K$ and $1.0$ atm pressure, when $5.4 \,g$ of aluminium and $500\, mL$ of $50 \,M$ sulfuric acid are combined for the reaction?
(Use molar mass of aluminium as $27.0 \,g\,mol ^{-1}, R=0.082 \,atm$ $L\,mol ^{-1} K^{-1}$

Answer 1

Correct Answer: 6.15

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Mole of Al taken = \(\frac{5.4 }{ 27}\) = 0.2

Mole of H₂SO₄ taken = \(\frac{{50 \times 5}}{{1000}} = 0.25\)

As \(\frac{0.2}{2} > \frac{0.25}{3}, \text{H}_2\text{SO}_4\) is limiting reagent

Now, moles of H₂ formed = \(\frac{3}{3} \times 0.25 = 0.25\)

Therefore Volume =\(0.25 \times 0.082 \times \frac{300}{1} = \frac{24.6}{4} = 6.15 \, \text{L}\)