15 g of a mixture of zinc and copper was treated with an excess of hydrochloric acid while 4.48 liters

15 g of a mixture of zinc and copper was treated with an excess of hydrochloric acid while 4.48 liters of gas were released, determine the composition of the mixture.

Given:
m mixture (Zn, Cu) = 15 g
V (gas) = 4.48 l

Find:
ω (Zn) -?
ω (Cu) -?

Solution:
1) Zn + 2HCl => ZnCl2 + H2 ↑;
Cu + HCl – the reaction will not proceed;
2) M (Zn) = Mr (Zn) = Ar (Zn) = 65 g / mol;
3) n (H2) = V (H2) / Vm = 4.48 / 22.4 = 0.2 mol;
4) n (Zn) = n (H2) = 0.2 mol;
5) m (Zn) = n (Zn) * M (Zn) = 0.2 * 65 = 13 g;
6) ω (Zn) = m (Zn) * 100% / m mixture (Zn, Cu) = 13 * 100% / 15 = 86.67%;
7) ω (Cu) = 100% – ω (Zn) = 100% – 86.67% = 13.33%.

Answer: The mass fraction of Zn is 86.67%; Cu – 13.33%.