The mass of the precipitate obtained by the interaction of 1.5 mol of copper sulfate with sodium hydroxide.

CuSO4 + 2NaOH = Cu (OH) 2 + Na2SO4 – ion exchange, copper hydroxide precipitate is precipitated;
Calculations using the formulas: M Cu (OH) 2 = 97.5 g / mol.
Based on the data of the problem, Y (CuSO4) = 1.5 mol, Y Cu (OH) 2 = 1.5 mol, since the amount of substances is 1 mol.
We find the mass of the sediment:
m Cu (OH) 2 = Y * M = 1.5 * 97.5 = 146.25 g

Answer: obtained copper hydroxide weighing 146.25 g