The reaction between aluminium and caustic soda is $\underset{2 \times 27 = 54\,\mathrm{g}}{2\mathrm{Al}}$ + 2NaOH + $2\mathrm{H}_2\mathrm{O}$ → $2\mathrm{NaAlO}_2 + \underset{3 \times 22.4\ \mathrm{L}\ \text{at STP}}{3\mathrm{H}_2}$ ∴ 54 g of Al produces $\mathrm{H}_2$ at S.T.P. = 3 × 22.4 L 0.15 g of Al will produce $\mathrm{H}_2$ at S.T.P. = $\frac{3 \times 22.4}{54}$ × 0.15 = 0.186 L At STP $P_1$ = 1 atm $V_1$ = 0.186 L $T_1$ = 273 K Given conditions $P_2$ = 1 bar = 0.987 atm $V_2$ = ? $T_2$ = 273 + 20 = 293 K Applying ideal gas equation, $\frac{P_1V_1}{T_1}$ = $\frac{P_2V_2}{T_2}$ $\frac{1 \times 0.186}{273}$ = $\frac{0.987 \times V_2}{293}$ ⇒ $V_2$ = $\frac{293}{0.987} \times \frac{1 \times 0.186}{273}$ = 0.2030 L = 203 mL