Answer is: (3 24.31) g of Mg produce (3 22.4) L of H2 gas.
Balanced chemical reaction:
3Mg(s) + 2H₃PO₄(aq) → Mg₃(PO₄)₂(s) + 3H₂(g).
From chemical reaction: n(Mg) : n(H₂) = 3 : 3 (1 : 1).
n(Mg) = n(H₂).
m(Mg) = 24.31 g; mass of magnesium.
n(Mg) = m(Mg) ÷ M(Mg).
n(Mg) = 24.31 g ÷ 24.31 g/mol.
n(Mg) = 1 mol; amount of the magnesium.
V(H₂) = n(H₂) · Vm.
V(H₂) = 1 mol · 22.4 L/mol.
V(H₂) = 22.4 L; volume of the hydrogen gas.
So, 3·24.31 g of magnesium (3 mol) produce 3·22.4 liters of hydrogen.
