Find the mass of the precipitate that forms when 4 mol of sodium hydroxide interacts with an excess of copper (II) sulfate solution.

2NaOH + CuSO4 = Na2SO4 + Cu (OH) 2
since n sodium hydroxide is 2 times more than n copper hydroxide
m (Cu (OH) 2) = M * n = 97.5 * 2 = 195 g
Answer: 195 grams



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