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In the reaction,
$2 Al (s) + 6 HCl (a q) ⟶ 6 {Cl}^{-} (a q) + 3 H_{2}$ .

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6 L HCl(aq) is consumed for every 3 L $H_{2}$ produced.

$33.6 L H_{2} (g)$ is produced regardless temperature and pressure for every moles that reacts.

$67.2 L H_{2} (g)$ at 1 atm, 273 K is produced for every mole Al that reacts.

$11.2 L H_{2} (g)$ at 1 atm, 273 K is produced for every mole HCI(aq) consumed

answer is D.

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Detailed Solution

For $6 mol HCl$ , $3 mol$ of $H_{2}$ gas is produced

At STP $1 mol$ of every gas has 22.4 L

so $1 mol (22.4 L)$ , will produce $3 / 6 * 22.4 = 1 / 2 * 22.4 = 11.2 L$ of $H_{2}$ gas.

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