$H_{2}$ evolved at STP on complete reaction of 27 g of aluminium with excess of aqueous NaOH would be

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22.4 litres

44.8 litres

67.2 litres

33.6 litres

answer is D.

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Detailed Solution

$\begin{array}{l} H_{2} O + \underset{27 g}{Al} + NaOH \to {NaAlO}_{2} + \frac{3}{2} H_{2} \\ volume of H_{2} evolved = \frac{3}{2} \times 22 .4 = 33 .6 L \end{array}$

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