In the reaction:
$2 A l (s) + 6 H C l (a q) \to 2 A l^{3 +} (a q) + 6 C l^{-} (a q) + 3 H_{2} (g)$ Calculate the volume of $H_{2}$ gas (L) produced at $S T P$ (atm) for every 3 mol of $H C l$ consumed.

33.6 L

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44.8 L

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67.2 L

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22.4 L

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Solution

The correct option is A $33.6 L$
According to the reaction,
6 mol of $H C l$ is consumed to produce 3 mol of $H_{2}$ gas, i.e. $3 \times 22.4 L$ of $H_{2}$ gas
3 mol of $H C l$ will produce = $\frac{3 \times 22.4}{6} \times 3 L = 33.6 L$

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2 A l (s) + 6 H C l (a q) \to 2 A l^{3 +} (a q) + 6 C l^{-} (a q) + 3 H_{2} (g)

H_{2}

S T P

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2 A l (s) + 6 H C l (a q) \to 2 A l^{3 +} (a q .) + 6 C l^{-} (a q .) + 3 H_{2} (g)

2 A l (s) + 6 H C l (a q) \to 2 A l^{3 +} (a q) + 6 C l^{-} (a q) + 3 H_{2} (g)

2 A l (s) + 6 H C l (a q) \to 2 {A l}^{3 +} (a q) + 6 {C l}^{-} (a q) + 3 H_{2} (g)

2 A l (s) + 6 H C l (a q) \to 2 {A l}^{3 +} (a q) + 6 {C l}^{-} (a q) + 3 H_{2} (g)

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