A solution containing 24 g of copper (II) sulfate was added to 100 g of a solution with a mass fraction of sodium

A solution containing 24 g of copper (II) sulfate was added to 100 g of a solution with a mass fraction of sodium hydroxide of 8%. The resulting precipitate was filtered off, dried and calcined. The solid residue obtained after calcination had a mass of 7.6 g. Calculate the yield of the reaction product.

Given:
m solution (NaOH) = 100 g
ω (NaOH) = 8%
m (CuSO4) = 24 g
m pract. (sediment) = 7.6 g

To find:
ω out. -?

Solution:
1) 2NaOH + CuSO4 => Cu (OH) 2 ↓ + Na2SO4;
Cu (OH) 2 = (tOC) => CuO + H2O;
2) M (NaOH) = Mr (NaOH) = Ar (Na) + Ar (O) + Ar (H) = 23 + 16 + 1 = 40 g / mol;
M (CuSO4) = Mr (CuSO4) = Ar (Cu) + Ar (S) + Ar (O) * 4 = 64 + 32 + 16 * 4 = 160 g / mol;
M (CuO) = Mr (CuO) = Ar (Cu) + Ar (O) = 64 + 16 = 80 g / mol;
3) m (NaOH) = ω (NaOH) * m solution (NaOH) / 100% = 8% * 100/100% = 8 g;
4) n (NaOH) = m (NaOH) / M (NaOH) = 8/40 = 0.2 mol;
5) n (CuSO4) = m (CuSO4) / M (CuSO4) = 24/160 = 0.15 mol;
6) The amount of NaOH substance is in short supply;
7) n (Cu (OH) 2) = n (NaOH) / 2 = 0.2 / 2 = 0.1 mol;
8) n (CuO) = n (Cu (OH) 2) = 0.1 mol;
9) m theor. (CuO) = n (CuO) * M (CuO) = 0.1 * 80 = 8 g;
10) ω out. = m practical (CuO) * 100% / m theory. (CuO) = 7.6 * 100% / 8 = 95%.

Answer: The yield of the reaction product is 95%.