How many grams of precipitate is formed when 160 grams of copper sulfate 2 interacts with potassium hydroxide?

To solve, we write down data on the condition of the problem:

m = 160 g X g -?

CuSO4 + 2KOH = Cu (OH) 2 + K2SO4 – ion exchange, copper hydroxide is released in the sediment.
We make calculations using the formulas:
M (CuSO4) = 159.5 g / mol;

M Cu (OH) 2 = 97.5 g / mol;

Y (CuSO4) = m / M = 160 / 159.5 = 1 mol;

Y Cu (OH) 2 = 1 mol since the amount of substances is 1 mol.

3. Find the mass of the sediment:

m Cu (OH) 2 = Y * M = 1 * 97.5 = 97.5 g

Answer: the mass of copper hydroxide is 97.5 g