An excess of copper sulfate was added to 120 g of sodium hydroxide solution with a mass fraction of alkali of 10% (2).

An excess of copper sulfate was added to 120 g of sodium hydroxide solution with a mass fraction of alkali of 10% (2). Determine the mass of precipitated sediment.

2NaOH + CuSO4 = Na2SO4 + Cu (OH) 2
m NaOH substances = m p-pa × w (NaOH) / 100% = 120 g × 0.1 = 12g
n (NaOH) = m (NaOH) in-va / M (NaOH) = 12 g / 40 g / mol = 0.3 mol
0.3 / 2 = n (Cu (OH) 2) / 1
n Cu (OH) 2 = 0.15 mol.
m Cu (OH) 2 = M (Cu (OH) 2) × n (Cu (OH) 2) = 98 g / mol × 0.15 mol = 14.7 g
Answer: 14.7 g