Calculate the mass of the precipitate obtained by the interaction of a salt solution in which 16

Calculate the mass of the precipitate obtained by the interaction of a salt solution in which 16 g of copper (II) sulfate, with an excess of sodium hydroxide solution.

Let’s write the reaction equation:

CuSO4 + 2NaOH = Cu (OH) 2 ↓ + Na2SO4

Let’s find the amount of copper (II) sulfate substance:

v (CuSO4) = m (CuSO4) / M (CuSO4) = 16/160 = 0.1 (mol).

According to the reaction equation, from 1 mol of CuSO4, 1 mol of Cu (OH) 2 is formed, therefore:

v (Cu (OH) 2) = v (CuSO4) = 0.1 (mol).

Thus, the mass of the formed copper (II) hydroxide precipitate is:

m (Cu (OH) 2) = v (Cu (OH) 2) * M (Cu (OH) 2) = 0.1 * 98 = 9.8 (g).

Answer: 9.8 g.