A sample of copper (II) oxide weighing 40 g was treated with a 10% sulfuric acid solution weighing 196 g

A sample of copper (II) oxide weighing 40 g was treated with a 10% sulfuric acid solution weighing 196 g. Calculate the mass of the precipitate in the reaction mixture.

Given:
m (CuO) = 40 g
ω (H2SO4) = 10%
m solution (H2SO4) = 196 g

Find:
m (draft) -?

Solution:
1) CuO + H2SO4 => CuSO4 + H2O;
2) n (CuO) = m / M = 40/80 = 0.5 mol;
3) m (H2SO4) = ω * m solution / 100% = 10% * 196/100% = 19.6 g;
4) n (H2SO4) = m / M = 19.6 / 98 = 0.2 mol;
5) n react. (CuO) = n (H2SO4) = 0.2 mol;
6) n rest. (CuO) = n (CuO) – n reag. (CuO) = 0.5 – 0.2 = 0.3 mol;
7) m rest. (CuO) = n rest. * M = 0.3 * 80 = 24 g.

Answer: The mass of CuO is 24 g.