Zinc weighing 130 g reacted with a hydrochloric acid solution to form 25 g of zinc chloride

Zinc weighing 130 g reacted with a hydrochloric acid solution to form 25 g of zinc chloride. Calculate the mass fraction of the zinc chloride yield.

Given:
m (Zn) = 130 g
m pract. (ZnCl2) = 25 g

To find:
η (ZnCl2) -?

Decision:
1) Zn + 2HCl => ZnCl2 + H2 ↑;
2) M (Zn) = Mr (Zn) = Ar (Zn) = 65 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 theory. (ZnCl2) = n (Zn) = 2 mol;
5) m theor. (ZnCl2) = n theory. (ZnCl2) * M (ZnCl2) = 2 * 136 = 272 g;
6) η (ZnCl2) = m practical. (ZnCl2) * 100% / m theor. (ZnCl2) = 25 * 100% / 272 = 9.2%.

Answer: The ZnCl2 yield is 9.2%.