In the reaction, 2Al(s) + 6HCl(aq) 2Al3+(aq) + 6Cl-(aq) + 3H2(g)
- 12 L HCl(aq) is consumed for every 6 L H2(g) produced.
- 11.2 L H2(g) at STP is produced for every mole of HCl consumed.
- 33.6 L H2(g) is produced regardless of temperature and pressure for every mole of Al that reacts.
- 67.2 L H2(g) at STP is produced for every mole of Al that reacts.
Answer
(2) 11.2 L H2(g) at STP is produced for every mole of HCl consumed.
Solution
From the equation: 2 mol Al ≡ 6 mol HCl ≡ 3 mol H2
(1) HCl is in aqueous solution, not a gas, so its volume is not related to the volume of H2 by the mole ratio. Incorrect.
(2) 6 mol HCl give 3 mol H2 1 mol HCl gives 0.5 mol H2
volume at STP = 0.5 22.4 = 11.2 L. Correct.
(3) The volume of a gas depends on temperature and pressure. Incorrect.
(4) 2 mol Al give 3 mol H2 1 mol Al gives 1.5 mol H2 = 1.5 22.4 = 33.6 L at STP, not 67.2 L. Incorrect.
Answer: (2)