An excess of potassium hydroxide solution was added to a 376 solution with a mass fraction of copper (2) nitrate of 7.5%.

An excess of potassium hydroxide solution was added to a 376 solution with a mass fraction of copper (2) nitrate of 7.5%. Determine the mass of the precipitate.

Cu (NO3) 2 + 2KOH = Cu (OH) 2 + 2KNO3
1) omega (W) = (m (substance): m (solution)) x100% 7,5 = (m (Cu (NO3) 2): 376) x100% => m (Cu (NO3) 2) = ( 376×7.5): 100 = 28.2 (gr.)
2) n (Cu (NO3) 2) = 28.2: (16×6 + 14×2 + 64) = 0.15 (mol.)
3) x n (Cu (OH) 2) = (0.15×2): 1 = 0.3 (mol.)
4) m (Cu (OH) 2) = 0.3x (64 + 16×3 + 2×1) = 34.2 (gr.)
Answer: the mass of the sediment is Cu (OH) 2 = 34.2 g.