How many grams of precipitate is formed when copper sulfate reacts with 500 g of 24% sodium hydroxide solution?

1.Let’s find the mass of sodium hydroxide in the solution.

500 g – 100%,

X = 24%,

X = (500 g x 24%): 100%,

X = 120g.

2.Let’s find the amount of sodium hydroxide substance by the formula:

n = m: M,

M (NaOH) = 23 + 16 + 1 = 40 g / mol.

n = 120 g: 40 g / mol = 3 mol.

CuSO4 + 2NaOH = Na2SO4 + Cu (OH) 2.

According to the reaction equation, there is 1 mole of copper hydroxide for 2 mol of sodium hydroxide. The substances are in quantitative ratios of 2: 1. The amount of copper substance is 2 times less than the amount of sodium hydroxide substance.

n (Cu (OH) 2) = 3: 2 = 1.5 mol.

3.Let’s find the mass of copper hydroxide by the formula: m = n M.

M (Cu (OH) 2) = 64 + 2 (16 + 1) = 64 + 34 = 98 g / mol.

m = 1.5 mol × 98 g / mol = 147 g.

Answer: 147 g.