A solution containing 10 g of sodium hydroxide was added to a solution of copper (II) sulfate

A solution containing 10 g of sodium hydroxide was added to a solution of copper (II) sulfate. Calculate the mass of the precipitate formed.

Given:
m (NaOH) = 10 g

To find:
m (draft) -?

Decision:
1) CuSO4 + 2NaOH => Cu (OH) 2 ↓ + Na2SO4;
2) n (NaOH) = m (NaOH) / M (NaOH) = 10/40 = 0.25 mol;
3) n (Cu (OH) 2) = n (NaOH) / 2 = 0.25 / 2 = 0.125 mol;
4) m (Cu (OH) 2) = n (Cu (OH) 2) * M (Cu (OH) 2) = 0.125 * 98 = 12.25 g.

Answer: The mass of Cu (OH) 2 is 12.25 g.