What volume of H2(g) is produced when 6.70 g of Al(s) reacts at STP? 2Al(s)+6HCl(aq) ---->2AlCl3(aq)+3H2(g)?

Molar volume of H₂ at STP = 22.4 L/mol
Molar mass of Al = 27.0 g/mol

2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)
Molar ratio Al : H₂ = 2 : 3

Method 1:
No. of moles of Al reacted = (6.70 g) / (27.0 g/mol) = 0.2481 mol
No. of moles of H₂ produced = (0.2481 mol) × (3/2) = 0.3722 mol
Volume of H₂ produced at STP = (0.3722 mol) × (22.4 L/mol) = 8.34 L

Method 2:
(6.70 g Al) × (1 mol Al / 27.0 g Al) × (3 mol H₂ / 2 mol Al) × (22.4 L H₂ / 1 mol H₂)
= 8.34 L H₂ produced