Find the mass of the precipitate that forms during the interactions of 320 g of copper sulfate, the yield is 80%

Find the mass of the precipitate that forms during the interactions of 320 g of copper sulfate, the yield is 80%, How much of the precipitate is copper hydroxide

1. Let’s write down the reaction equation:

CuSO4 + 2KOH = Cu (OH) 2 + K2SO4.

2. Find the amount of reacted copper sulfate:

n (CuSO4) = m (CuSO4) / M (CuSO4) = 320 g / 160 g / mol = 2 mol.

3. According to the reaction equation, we find the theoretical amount of copper hydroxide, and at the end its mass:

ntheor (Cu (OH) 2) = n (CuSO4) = 2 mol.

nprak (Cu (OH) 2) = ntheor (Cu (OH) 2) * η (Cu (OH) 2) / 100% = 2 mol * 80% / 100% = 1.6 mol.

m (Cu (OH) 2) = nprak (Cu (OH) 2) * M (Cu (OH) 2) = 1.6 mol * 98 g / mol = 156.8 g.

Answer: m (Cu (OH) 2) = 156.8 g.