Question

What volume of H2(g) is produced when 8.80 g of Al(s) reacts at STP?

Answer 1

Question

What volume of $\mathrm{H}_2(\mathrm{g})$ is produced when $8.80\mathrm{g}$ of Al(s) reacts at STP?

Solution

In all reactions of Al(s) resulting in $\mathrm{H}_{2}$ formation the molar ratio Al/H₂ is always 2/3, e.g.

$$2 \mathrm{Al} + 6 \mathrm{H}_2\mathrm{O} = 2 \mathrm{Al}(\mathrm{OH})_3 + 3 \mathrm{H}_2$$ $$2 \mathrm{Al} + 6 \mathrm{HCl} = 2 \mathrm{AlCl}_3 + 3 \mathrm{H}_2$$

Number of moles of Al:

$$n(\mathrm{Al}) = m(\mathrm{Al}) / M(\mathrm{Al}) = 8.80 / 26.98 = 0.33\ \mathrm{mol}$$ $$n(\mathrm{H}_2) = n(\mathrm{Al}) \cdot 2/3 = 0.33 \cdot 3/2 = 0.49\ \mathrm{mol}$$

At STP 1 mol of a gas occupies 22.4 L.

So, the volume of hydrogen produced is

$$V(\mathrm{H}_2) = n(\mathrm{H}_2) \cdot 22.4 = 0.49 \cdot 22.4 = 11.0\ \mathrm{L}$$

Answer: 11.0 L