How many grams of copper is formed during the reduction of copper (2) oxide with hydrogen

How many grams of copper is formed during the reduction of copper (2) oxide with hydrogen, if it was obtained by the interaction of 400 grams of 9.8% sulfuric acid and 300 grams of magnesium (20% impurity)?

Given:
m solution (H2SO4) = 400 g
ω (H2SO4) = 9.8%
m tech. (Mg) = 300 g
ω approx. = 20%

To find:
m (Cu) -?

Decision:
1) H2SO4 + Mg => MgSO4 + H2 ↑;
2) m (H2SO4) = ω (H2SO4) * m solution (H2SO4) / 100% = 9.8% * 400/100% = 39.2 g;
3) n (H2SO4) = m (H2SO4) / M (H2SO4) = 39.2 / 98 = 0.4 mol;
4) ω (Mg) = 100% – ω approx. = 100% – 20% = 80%;
5) m clean. (Mg) = ω (Mg) * m tech. (Mg) / 100% = 80% * 300/100% = 240 g;
6) n (Mg) = m pure. (Mg) / M (Mg) = 240/24 = 10 mol;
7) n (H2) = n (H2SO4) = 0.4 mol;
8) CuO + H2 => Cu + H2O;
9) n (Cu) = n (H2) = 0.4 mol;
10) m (Cu) = n (Cu) * M (Cu) = 0.4 * 64 = 25.6 g.

Answer: The mass of Cu is 25.6 g.