Question

The drain cleaner called 'Drainex' contains small bits of aluminium which react with caustic soda to produce hydrogen. What volume of hydrogen at 27°C and 0.831 bar will be released when 0.15 g of aluminium reacts? (Al = 27)

• A. 250 ml
• B. 500 ml
• C. 150 ml
• D. 125 ml

Answer 1

Answer: (A)

250 ml

Answer 2

1. Balanced equation: $2Al + 2NaOH + 2H_2O \rightarrow 2NaAlO_2 + 3H_2$.
2. Moles of Al: $n_{Al} = \frac{0.15 \text{ g}}{27 \text{ g/mol}} = \frac{1}{180} \text{ mol}$.
3. Moles of $H_2$: From stoichiometry, 2 moles of Al produce 3 moles of $H_2$. So, $n_{H_2} = \frac{3}{2} \times n_{Al} = \frac{3}{2} \times \frac{1}{180} = \frac{1}{120} \text{ mol}$.
4. Ideal Gas Law ($PV = nRT$): Given: $P = 0.831 \text{ bar}$, $T = 27^\circ C = 300 \text{ K}$, $n = \frac{1}{120} \text{ mol}$, $R = 0.0831 \text{ L bar/mol K}$. $V = \frac{nRT}{P} = \frac{(\frac{1}{120}) \times (0.0831) \times (300)}{0.831} \text{ L} = 0.25 \text{ L}$.
5. Convert to ml: $0.25 \text{ L} \times 1000 \text{ ml/L} = 250 \text{ ml}$.