Answer: The reaction produces 2.93 g H₂.
M_r: 133.34 2.016
2Al + 6HCl → 2AlCl₃ + 3H₂
Moles of AlCl₃ = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃
Moles of H₂ = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂
Mass of H₂ = 1.451 mol H₂ × (2.016 g H₂/1 mol H₂) = 2.93 g H₂
Explanation:
