Mixed a portion of a 10% sodium hydroxide solution weighing 80 g and a portion of a 40% solution of copper (II)

Mixed a portion of a 10% sodium hydroxide solution weighing 80 g and a portion of a 40% solution of copper (II) sulfate weighing 80 g. Determine the mass of the precipitate formed and the mass fraction of the salt formed in the solution after the reaction.

Given:
ω (NaOH) = 10%
m solution (NaOH) = 80 g
ω (CuSO4) = 40%
m solution (CuSO4) = 80 g

Find:
m (draft) -?
ω (salt) -?

Solution:
1) 2NaOH + CuSO4 => Cu (OH) 2 ↓ + Na2SO4;
2) m (NaOH) = ω * m solution / 100% = 10% * 80/100% = 8 g;
3) n (NaOH) = m / M = 8/40 = 0.2 mol;
4) m (CuSO4) = ω * m solution / 100% = 40% * 80/100% = 32 g;
5) n (CuSO4) = m / M = 32/160 = 0.2 mol;
6) n (Cu (OH) 2) = n (NaOH) / 2 = 0.2 / 2 = 0.1 mol;
7) m (Cu (OH) 2) = n * M = 0.1 * 98 = 9.8 g;
8) n (Na2SO4) = n (NaOH) / 2 = 0.2 / 2 = 0.1 mol;
9) m (Na2SO4) = n * M = 0.1 * 142 = 14.2 g;
10) m2 solution = m solution (NaOH) + m solution (CuSO4) – m (Cu (OH) 2) = 80 + 80 – 9.8 = 150.2 g;
11) ω (Na2SO4) = m * 100% / m2 solution = 14.2 * 100% / 150.2 = 9.5%.

Answer: The mass of Cu (OH) 2 is 9.8 g; mass fraction of Na2SO4 – 9.5%.