When treating 15 g of a mixture of aluminum, silicon, copper and copper (II) oxide hydrochloric acid, 5.04 liters of gas were released

When treating 15 g of a mixture of aluminum, silicon, copper and copper (II) oxide hydrochloric acid, 5.04 liters of gas were released, and the mass of the remaining sediment was 4.2 g. Determine the mass fraction of copper (II) oxide in the initial mixture.

2Al + 6HCl → 2AlCl3 + 3H2 ↑
Si + HCl ≠
Cu + HCl ≠
CuO + HCl → CuCl2 + H2O
m (mixture) = 15 g
V (H2) = 5.04 L
m (sediment) = 4.2 g
n (H2) = 5.04 / 22.4 = 0.225 mol
⇒ n (Al) = 0.225: 3 x 2 = 0.15 mol
m (Al) = 0.15 x 27 = 4.05 g
m (Al + CuO) = 15 – 4.2 = 10.8 g
⇒ m (CuO) = 10.8 – 4.05 = 6.75 g
ω (CuO) = 6.75 / 15 x 100% = 45%
Answer: the mass fraction of copper oxide in the initial mixture is 45%.

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