To 150 g of ammonium chloride solution, 240 g of 28% potassium hydroxide solution was added

| May 25, 2021 | education

To 150 g of ammonium chloride solution, 240 g of 28% potassium hydroxide solution was added and the resulting solution was heated until gas evolution ceased. To completely neutralize the unreacted alkali, 245 g of a 16% sulfuric acid solution was added. Determine the mass fraction of ammonium chloride in the original solution.

Let’s find the mass of KOH in solution.

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

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

m (KOH) = (240 g × 28%): 100% = 67.2 g.

Find the mass of H2SO4 in solution.

m (H2SO4) = (245 g × 16%): 100% = 39.2 g.

M (H2SO4) = 98 g / mol.

n = 39.2 g: 98 g / mol = 0.4 mol.

NH4Cl + KOH = NH4OH + KCl.

2KOH + H2SO4 = K2SO4 + 2H2O.

For 1 mol of H2SO4, there are 2 mol of KOH. The substances are in quantitative ratios of 1: 2. The amount of KOH substance is 2 times greater than the amount of H2SO4 substance.

n (KOH) = 2n (H2SO4) = 0.4 × 2 = 0.8 mol.

Let’s find the mass of KOH.

M (KOH) = 56 g / mol.

m = n × M.

m = 56 g / mol × 0.8 mol = 44.8 g.

Let us find the mass of KOH that has reacted with NH4Cl.

m (KOH) 67.2 g – 44.8 g = 22.4 g.

n (KOH) = 2.4: 56 g / mol = 0.4 mol.

n (KOH) = n (NH4Cl) = 0.4 mol.

m (NH4Cl) = 0.4 mol × 53.5 g / mol = 21.4 g.

W = (21.4g: 150g) × 100% = 14.27%.

Answer: 14.27%.



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