The electrolysis of an aqueous solution of copper sulfate with w (cuso4) = 15% was continued until

The electrolysis of an aqueous solution of copper sulfate with w (cuso4) = 15% was continued until the mass fraction of copper sulfate was 7%. Find the masses of all substances released on the electrodes, and the mass fraction of acid in the final solution, if the mass of the initial solution of copper sulfate is 500 g.

Given:
m out. solution (CuSO4) = 500 g
ω ref. (CuSO4) = 15%
ω end. (CuSO4) = 7%

To find:
m end. (in-in) -?
ω end. (acid) -?

Decision:
1) 2CuSO4 + 2H2O = (electric current) => 2Cu + 2H2SO4 + O2;
2) Let n react. (CuSO4) = x mol;
3) m react. (CuSO4) = n reag. (CuSO4) * M (CuSO4) = (160x) g;
4) n (Cu) = n reag. (CuSO4) = x mol;
5) m (Cu) = n (Cu) * M (Cu) = (64x) g;
6) n (O2) = n reag. (CuSO4) / 2 = (x / 2) mol;
7) m (O2) = n (O2) * M (O2) = ((x / 2) * 32) = (16x) g;
8) m end. solution = m ref. solution (CuSO4) – m (Cu) – m (O2) = 500 – 64x – 16x = (500 – 48x) g;
9) m out. (CuSO4) = ω ref. (CuSO4) * m ref. solution (CuSO4) / 100% = 15% * 500/100% = 75 g;
10) m end. (CuSO4) = m ref. (CuSO4) – m reag. (CuSO4) = (75 – 160x) g;
11) ω end. (CuSO4) = m end. (CuSO4) * 100% / m end. solution;
7% = (75 – 160x) * 100% / (500 – 48x);
35 – 3.36x = 75 – 160x;
156.64x = 40;
x = 0.26;
n react. (CuSO4) = x = 0.26 mol;
12) m (Cu) = 64x = 64 * 0.26 = 16.64 g;
13) m (O2) = 16x = 16 * 0.26 = 4.16 g;
14) n (H2SO4) = n reag. (CuSO4) = 0.26 mol;
15) m (H2SO4) = n (H2SO4) * M (H2SO4) = 0.26 * 98 = 25.48 g;
16) m end. solution = 500 – 48x = 500 – 48 * 0.26 = 487.52 g;
17) ω (H2SO4) = m (H2SO4) * 100% / m end. p-ra = 25.48 * 100% / 487.52 = 5.23%.

Answer: The mass of Cu is 16.64 g; O2 4.16 g; mass fraction of H2SO4 – 5.23%.