100 mL of 5 M $NaOH$ solution (density 1.2 g/ml) added to 200 mL of another $NaOH$ solution which as a density of 1.5 g/ml and contains 20 mass percent of $NaOH$ . What will be the volume of the gas (at STP) in litres liberated when aluminium reacts with this (final) solution? [Give answer excluding decimal places] The reaction is $Al + NaOH + H_{2} O ⟶ {NaAlO}_{2} + H_{2}$ (At. wt. $Na = 23, O = 16, H = 1)$

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answer is 68.

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Detailed Solution

$Al + NaOH + H_{2} O \to {NaAlO}_{2} + \frac{3}{2} H_{2}$

$[0.1 \times 5 + 200 \times \frac{1.5}{40} \times \frac{20}{100}] = 2 moles$

$V_{H_{2}} = 2 \times 22.7 \times \frac{3}{2} = 68.1 L$ .

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