When aluminum reacts with concentrated hydrochloric acid, hydrogen gas is produced. 2Al(s)+6HCl(aq)⟶2AlCl3(aq)+3H2(g) What mass of Al(s) is required to produce 717.5 mL of H2(g) at STP?

Answer 1

2Al(s) + 6HCl(aq) ==> 2AlCl₃(aq) + 3H₂(g) .. balanced equation

At STP, 1 mole of any ideal gas = 22.4 L

moles H₂(g) = 717.5 ml x 1 L / 1000 ml x 1 mol / 22.4 L = 0.03203 mols H₂(g)

Now use the stoichiometry of the balanced equation along with dimensional analysis (unit factor):

0.03203 mols H₂ x 2 mol Al / 3 mol H₂ = 0.02135 mols Al

atomic mass Al = 26.98 g / mol

mass of Al = 002135 mols Al x 26..98 g / mol = 0.5761 g Al