Zinc weighing 130 g reacts with hydrochloric acid in the amount of 4 mol according to the following reaction equation Zn

Zinc weighing 130 g reacts with hydrochloric acid in the amount of 4 mol according to the following reaction equation Zn + 2HCl = ZnCl2 + H2. This releases 4 grams of hydrogen. Find the mass of zinc chloride formed.

Given:
m (Zn) = 130 g
n (HCl) = 4 mol
m (H2) = 4 g

To find:
m (ZnCl2) -?

Decision:
1) Zn + 2HCl => ZnCl2 + H2 ↑;
2) M (Zn) = Ar (Zn) = 65 g / mol;
M (H2) = Mr (H2) = Ar (H) * N (H) = 1 * 2 = 2 g / mol;
M (ZnCl2) = Mr (ZnCl2) = Ar (Zn) * N (Zn) + Ar (Cl) * N (Cl) = 65 * 1 + 35.5 * 2 = 136 g / mol;
3) n (Zn) = m (Zn) / M (Zn) = 130/65 = 2 mol;
4) n (H2) = m (H2) / M (H2) = 4/2 = 2 mol;
5) n (ZnCl2) = n (Zn) = n (H2) = 2 mol;
6) m (ZnCl2) = n (ZnCl2) * M (ZnCl2) = 2 * 136 = 272 g.

Answer: The mass of ZnCl2 is 272 g.