10.04 g of a mixture of zinc and nickel (mass fraction of zinc in the mixture is 65%) was dissolved in 20%

10.04 g of a mixture of zinc and nickel (mass fraction of zinc in the mixture is 65%) was dissolved in 20% hydrochloric acid (density 1.1 g / ml). calculate the volume of the acid solution.

Given:
m mixture (Zn, Ni) = 10.04 g
ω (Zn) = 65%
ω (HCl) = 20%
ρ solution (HCl) = 1.1 g / ml

To find:
V (HCl) -?

Decision:
1) Zn + 2HCl => ZnCl2 + H2;
Ni + 2HCl => NiCl2 + H2;
2) m (Zn) = ω (Zn) * m mixture (Zn, Ni) / 100% = 65% * 10.04 / 100% = 6.526 g;
3) n (Zn) = m (Zn) / M (Zn) = 6.526 / 65 = 0.1 mol;
4) n1 (HCl) = n (Zn) * 2 = 0.1 * 2 = 0.2 mol;
5) m (Ni) = m mixture (Zn, Ni) – m (Zn) = 10.04 – 6.526 = 3.514 g;
6) n (Ni) = m (Ni) / M (Ni) = 3.514 / 59 = 0.06 mol;
7) n2 (HCl) = n (Ni) * 2 = 0.06 * 2 = 0.12 mol;
8) n total (HCl) = n1 (HCl) + n2 (HCl) = 0.2 + 0.12 = 0.32 mol;
9) m (HCl) = n total. (HCl) * M (HCl) = 0.32 * 36.5 = 11.68 g;
10) m solution (HCl) = m (HCl) * 100% / ω (HCl) = 11.68 * 100% / 20% = 58.4 g;
11) V solution (HCl) = m solution (HCl) / ρ solution (HCl) = 58.4 / 1.1 = 53.09 ml.

Answer: The volume of the HCl solution is 53.09 ml.