The drain cleaner,Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at $20^{\circ} C$ and $1 \, bar$ will be released when $0.15 \, g$ of aluminum reacts?

(D) The reaction of aluminum with caustic soda is given by:
$2Al(s) + 2NaOH(aq) + 2H_2O(l) \to 2NaAlO_2(aq) + 3H_2(g)$
From the stoichiometry,$2 \, mol$ of $Al$ $(54 \, g)$ produces $3 \, mol$ of $H_2$.
At $STP$ ($273.15 \, K$ and $1 \, bar$),$1 \, mol$ of an ideal gas occupies $22.7 \, L$ $(22700 \, mL)$.
Volume of $H_2$ at $STP$ produced by $0.15 \, g$ of $Al$:
$V_{STP} = \frac{3 \times 22700 \times 0.15}{54} \, mL = 189.17 \, mL$
Using the combined gas law $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$:
$P_1 = 1 \, bar, V_1 = 189.17 \, mL, T_1 = 273.15 \, K$
$P_2 = 1 \, bar, T_2 = 293.15 \, K$
$V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{1 \times 189.17 \times 293.15}{1 \times 273.15} \approx 203 \, mL$
Thus,$203 \, mL$ of dihydrogen is released.