Find the m precipitate precipitated by the interaction of 160 g of CuSO4 with an excess of NaOH solution, if the product yield is 80%?

Given: m (CuSO4) = 160g η = 80% NaOH – excess
Find: m (draft) -?
Decision.
CuSO4 + 2NaOH = Na2SO4 + Cu (OH) 2 Cu (OH) 2 – sediment, so you need to find its mass. n = m / M n (CuSO4) = 160/160 = 1 mol Since NaOH is in excess, CuSO4 is in short supply, so we calculate it using it: m = M * nm (Cu (OH) 2) = 97 * 1 = 97g – theoretical mass Product yield η = (m practical / m theoretical) * 100% => m practical = η * m theoretical / 100 m pr. (Cu (OH) 2) = 80 * 97/100 = 77.6 r
Answer: 77.6g.