What is the mass of the precipitate formed when an excess of sodium hydroxide

What is the mass of the precipitate formed when an excess of sodium hydroxide solution interacts with a solution containing 25 g of copper (II) sulfate.

Let’s find the amount of substance CuSO4 by the formula:

n = m: M.

M (CuSO4) = 160 g / mol.

n = 25 g: 160 g / mol = 0.156 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

CuSO4 + 2NaOH = Cu (OH) 2 ↓ + Na2SO4.

According to the reaction equation, there is 1 mol of Cu (OH) 2 for 1 mol of CuSO4. The substances are in quantitative ratios of 1: 1.

n (CuSO4) = n (Cu (OH) 2) = 0.156 mol.

Find the mass of Cu (OH) 2 by the formula:

m = n × M,

M (Cu (OH) 2) = 98 g / mol.

m = 0.156 mol × 98 g / mol = 15.29 g.

Answer: 15.29 g.