# To a 20% solution of acetic acid weighing 75 g was added magnesium oxide weighing 30 g. Calculate

**To a 20% solution of acetic acid weighing 75 g was added magnesium oxide weighing 30 g. Calculate the mass fraction of magnesium acetate in the solution after the reaction.**

W = m (substance): m (solution) × 100%,

m (substance) = (m (solution) × W): 100%.

m (CH3COOH) = (75 g × 20%): 100% = 1.5 g.

Find the amount of CH3COOH by the formula:

n = m: M.

M (CH3COOH) = 60 g / mol.

n = 1.5 g: 60 g / mol = 0.025 mol (deficiency).

Find the amount of MgO.

n = m: M.

M (MgO) = 40 g / mol.

n = 30 g: 40 g / mol = 0.75 mol (excess).

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

Mg + 2CH3COOH → (CH3COO) 2Mg + H2.

According to the reaction equation, there is 1 mol of (CH3COO) 2Mg per 2 mol of CH3COOH. Substances are in quantitative ratios of 2: 1.

The amount of (CH3COO) 2Mg substance will be 2 times less than the amount of CH3COOH.

n ((CH3COO) 2Mg) = ½ n (CH3COOH) = 0.025: 2 = 0.0125 mol.

Let’s find the mass (CH3COO) 2Mg by the formula:

m = n × M,

M (CH3CO) 2Mg) = 142 g / mol.

m = 0.0125 mol × 142 g / mol = 1.775 g.

W = m (substance): m (solution) × 100%,

m (solution) = 75 g + 30 g = 105 g.

W = (1.775 g: 105 g) × 100% = 16.9%.

Answer: 16.9%.