A mixture of copper (2) oxide and aluminum with a total mass of 15.2 g was ignited using a magnesium tape.

A mixture of copper (2) oxide and aluminum with a total mass of 15.2 g was ignited using a magnesium tape. After the end of the reaction, the resulting solid precipitate was partially dissolved in hydrochloric acid with the evolution of 6.72 liters of gas. Calculate the mass fractions (in%) of substances in the initial mixture. Take the relative atomic mass of copper to be 64.

Let’s compose the equations of the reactions that can occur in the mixture.

1) 3CuO + 2Al = Cu + Al2O3.

2) 2Al + 6HCl = 2AlCl3 + 3H2 ↑.

Let’s find the amount of the substance hydrogen.

n = V: Vn.

n (H2) = 6.72 L: 22.4 L / mol = 0.3 mol.

According to the reaction equation, there are 2 mol of aluminum for 3 mol of hydrogen. Substances react in quantitative ratios 3: 1. The amount of aluminum substance is 1.5 times less than the amount of hydrogen substance.

n (Al) = 0.3 mol: 1.5 = 0.2 mol.

Let’s find the mass of hydrogen.

m = n × M.

M (Al) = 27 g / mol.

m = 27 g / mol × 0.2 mol = 5.4 g.

The mass of the mixture is 15.2 g.

Find the rest of the mixture.

15.2 g – 5.4 g = 9.8 g.

Let’s find the mass of copper and aluminum, making the proportion according to the first equation.

3CuO + 2Al = Cu + Al2O3.

x is the mass of CuO,

mass of aluminum: 9.8 – x.

x: (3 × 80) = (9.8 – x): 2 × 27,

x / 240 = (9.8 – x) / 54,

x = 240 (9.8 – x) / 54,

x = 2352 – 240 x / 54,

54 x = 2352 – 240 x,

54 x + 240 x = 2352,

294 x = 2352,

X = 8.

CuO mass – 8 g.

w = (8g: 15.2) × 100% = 52.63%.

w (Al) = 100% – 52.63% = 47.37%.

Answer: 52.63%; 47.37%.