How many grams of precipitate is formed when 160 g of a 15% solution of CuSO4 interacts with the required amount of NaOH?

How many grams of precipitate is formed when 160 g of a 15% solution of CuSO4 interacts with the required amount of NaOH? Calculate the amount of sediment substance.

1. Determine how much salt in grams was in the solution. 160 * 0.15 = 24
2. Determine the number of moles of salt in the solution. The molar mass of CuSO4 is 159 g / mol. 24/159 = 0.151 mol
According to the reaction CuSO4 + 2NaOH = Cu (OH) 2
According to the ratio of the precipitate, the same mol is formed as the original salt that was in solution
Molar mass of Cu (OH) 2 = 97.56 g / mol
3 calculate the mass of the sediment
m = 0.151 * 97.56 = 14.72 g