Reaction: 2Al(s)+6HCl(aq)→2Al3+(aq)+6Cl−(aq)+3H2(g)
Checking each option:
(1) Per 1 mole HCl: From stoichiometry, 6 mol HCl produces 3 mol H₂, so 1 mol HCl produces 0.5 mol H₂ = 11.2 L at STP. TRUE
(2) Per 1 mole Al: 2 mol Al produces 3 mol H₂, so 1 mol Al produces 1.5 mol H₂ = 33.6 L at STP only, not regardless of T and P. FALSE
(3) Volume ratio at same T,P: 6 mol HCl : 3 mol H₂ = 2:1, so 12 L HCl : 6 L H₂. TRUE
(4) Per 1 mole Al: 1.5 mol H₂ = 33.6 L, not 67.2 L. FALSE
Option (1) is correct.