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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 formed, independent of temperature and pressure, for every mole of Al that reacts
4. 67.2 L of H₂(g) at STP is formed for every 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

2Al->3H2 means 1 mol Al gives 1.5 mol H2 = 33.6 L (not 67.2). The correct true statement is option 0: 6 HCl -> 3 H2, so 1 mol HCl gives 0.5 mol H2 = 11.2 L at STP.

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