How many grams of copper hydroxide is formed by the interaction of 10 g of CuSo4 and sodium hydroxide.

Let’s write the equation:
CuSO4 + 2NaOH = Cu (OH) 2 + Na2SO4 – during the ion exchange reaction, a copper hydroxide precipitate was obtained;
Let’s make calculations:
M (CuSO4) = 159.5 g / mol;
M Cu (OH) 2 = 97.5 g / mol.
Let’s calculate the amount of moles of copper sulfate, if the mass is known:
Y (CuSO4) = m / M = 10/159.5 = 0.06 mol;
According to the reaction equation, the amount of moles of copper sulfate, copper hydroxide is equal to 1 mole, which means that YСu (OH) 2 = 0.06 mol.
Determine by the formula:
m Cu (OH) 2 = Y * M = 0.06 * 97.5 = 5.85 g.
Answer: The mass of the copper hydroxide precipitate is 5.85 g.