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For the reaction 2Al(s) + 6HCl(aq) → 2Al³+(aq) + 6Cl^-(aq) + 3H2(g), which statement is correct?

1. 11.2 L of H2(g) at STP is formed for each mole of HCl(aq) used up.
2. 6 L of HCl(aq) is used for every 3 L of H2(g) formed.
3. 33.6 L of H2(g) is obtained independent of temperature and pressure for each mole of Al that reacts.
4. 67.2 L of H2(g) at STP is obtained for each mole of Al that reacts.

Correct answer: 11.2 L of H2(g) at STP is formed for each mole of HCl(aq) used up.

Solution

From the stoichiometry 6 HCl -> 3 H2, each mole of HCl yields 0.5 mol H2 = 0.5 x 22.4 = 11.2 L at STP. (1 mol Al gives 1.5 mol = 33.6 L H2, not 67.2 L, so that option is wrong.)

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