Aluminum reacts with sulfuric acid to form aluminum sulfate and hydrogen. What is the volume of hydrogen gas in liters $(L)$ produced at 300 $K$ and 1.0 $a t m$ pressure, when 5.4 $g$ of aluminum and 50.0 $m L$ of 5.0 $M$ sulfuric acid are combined for the reaction?

(Use molar mass of aluminum as $27.0 {g mol}^{- 1}, R = 0.082 {atm L mol}^{- 1} K^{- 1})$

6.150

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6.1

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6.15

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Solution

$2 A l + 3 H_{2} S O_{4} \to A l_{2} (S O_{4})_{3} + 3 H_{2} Moles of Al taken = \frac{5.4}{27} = 0.2 Mole of H_{2} S O_{4} taken = 50 \times 5 / 100 = 0.25 As 0.2 / 2 > 0.25 / 3, H_{2} S O_{4} is limiting reagent Now, moles of H_{2} formed = 3 / 3 \times 0.25 = 0.25$
Therefore, $Volume = 0.25 \times 0.082 \times 300 / 1 = 24.6 / 4 = 6.15 L$

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