What amount of copper (II) hydroxide substance was formed during the precipitation of 10 g of copper (II) sulfate solution

What amount of copper (II) hydroxide substance was formed during the precipitation of 10 g of copper (II) sulfate solution with a mass fraction of CUSO4 in it 5% by sodium hydroxide?

Reaction: 2NaOH + CuSO4 = Cu (OH) 2 + Na2SO4
Let’s find the amount of copper sulfate substance in its solution mv-va = w * mr-pa = 0.05 * 10 g = 0.5 g. Let’s find the amount of copper sulfate substance n = m / M = 0.5 g / 160 g / mol = 0.003125 mol. (M – molar mass, determined by Mendeleev’s table). According to the reaction, 1 mol of hydroxide is formed from 1 mol of copper sulfate, n (CuSO4) = n (Cu (OH) 2) = 0.003125 mol.
Answer: 0.003125 mol of copper hydroxide was formed.