15 g of a mixture of iron, aluminum, CuO was reduced with hydrogen. if the reaction mixture is treated with HNO3

15 g of a mixture of iron, aluminum, CuO was reduced with hydrogen. if the reaction mixture is treated with HNO3, then 2.24 liters of gas are released, when treated with hydrochloric acid, 6.72 liters of gas are released. Determine the quantitative composition of the mixture in grams.

Given:
m mixture = 15 g
V1 (gas) = ​​2.24 l
V2 (gas) = ​​6.72 l

To find:
m (Fe) -?
m (Al) -?
m (CuO) -?

Solution:
1) CuO + H2 => Cu + H2O;
Cu + 4HNO3 => Cu (NO3) 2 + 2NO2 ↑ + 2H2O;
Fe + 2HCl => FeCl2 + H2 ↑;
2Al + 6HCl => 2AlCl3 + 3H2 ↑;
2) n (NO2) = V1 (gas) / Vm = 2.24 / 22.4 = 0.1 mol;
3) n (CuO) = n (Cu) = n (NO2) / 2 = 0.1 / 2 = 0.05 mol;
4) m (CuO) = n (CuO) * M (CuO) = 0.05 * 80 = 4 g;
5) m (Fe, Al) = m mixture – m (CuO) = 15 – 4 = 11 g;
6) Let m (Fe) = (x) r, then m (Al) = (11 – x) r;
7) n (Fe) = m (Fe) / M (Fe) = (x / 56) mol;
8) n1 (H2) = n (Fe) = (x / 56) mol;
9) n (Al) = m (Al) / M (Al) = ((11 – x) / 27) mol;
10) n2 (H2) = n (Al) * 3/2 = ((11 – x) / 27) * 3/2 = ((11 – x) / 18) mol;
11) n total (H2) = V2 (gas) / Vm = 6.72 / 22.4 = 0.3 mol;
12) n total (H2) = n1 (H2) + n2 (H2);
0.3 = (x / 56) + ((11 – x) / 18);
x = 8.25;
13) m (Fe) = x = 8.25 g;
14) m (Al) = 11 – x = 11 – 8.25 = 2.75 g.

Answer: The mass of Fe is 8.25 g; Al 2.75 g; CuO – 4 g.