Question

what mass of hydrochloric acid reacts when 87.7 grams of aluminum dissolves in excess hydrochloric acid?

Answer 1

2Al + 6HCl = 2AlCl3 + 3H2

n = m/M

M (Al) = 26.9 g/mol

M (HCl) = 36.5 g/mol

n (Al) = 87.7/26.9 = 3.3 mol

n (HCl) = 6 · (n (Al)/2) = 6 · (3.3/2) = 9.8 mol

m (HCl) = n (HCl) · M (HCl) = 9.8 · 36.5 = 357 g