10 g of a mixture of copper and aluminum was treated with hydrochloric acid, while 6.72 liters of hydrogen

10 g of a mixture of copper and aluminum was treated with hydrochloric acid, while 6.72 liters of hydrogen were released. determine the composition of the mixture.

Given:
m mixture (Cu, Al) = 10 g
V (H2) = 6.72 L
Vm = 22.4 l / mol

Find:
ω (Cu) -?
ω (Al) -?

Solution:
1) Cu + HCl – the reaction will not proceed;
2Al + 6HCl => 2AlCl3 + 3H2 ↑;
2) n (H2) = V (H2) / Vm = 6.72 / 22.4 = 0.3 mol;
3) n (Al) = n (H2) * 2/3 = 0.3 * 2/3 = 0.2 mol;
4) m (Al) = n (Al) * M (Al) = 0.2 * 27 = 5.4 g;
5) ω (Al) = m (Al) * 100% / m mixture (Cu, Al) = 5.4 * 100% / 10 = 54%;
6) ω (Cu) = 100% – ω (Al) = 100% – 54% = 46%.

Answer: The mass fraction of Al is 54%; Cu – 46%.