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 50.0 mL of 5.0 M sulfuric acid are combined for the reaction? (nearest integer)
(Use molar mass of aluminium as $27.0 g {mol}^{- 1}, R = 0.08 atm L {mol}^{- 1} K^{- 1})$

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Detailed Solution

$2 Al + 3 H_{2} {SO}_{4} \to {Al}_{2} {({SO}_{4})}_{3} + 3 H_{2}$
Moles of Al takes $= \frac{5 .4}{27} = 0 .2$
Moles of H₂SO₄ taken $= \frac{50 \times 50}{1000} = 0 .25$
As $\frac{0 .2}{2} > \frac{0 .25}{3}, H_{2} {SO}_{4}$ is limiting reagent
Now, moles of H₂ formed $1 = \frac{3}{3} \times 0 .25 = 0 .25$
$∴$ Volume of H₂ gas formed $1 = \frac{nRT}{P} = \frac{0 .25 \times 0 .082 \times 300}{1} = 6 L$

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