By acting on a mixture of sodium chloride and potassium chloride with a total mass of 261 g with an excess

By acting on a mixture of sodium chloride and potassium chloride with a total mass of 261 g with an excess of concentrated sulfuric acid, 89.6 liters of hydrogen chloride were obtained. Determine the mass fraction of potassium chloride in the original mixture.

Let’s find the amount of substance H2.

n = V: Vn.

n = 89.6 L: 22.4 L / mol = 4 mol.

2NaCl + H2SO4 = 2NaCl + 2HCl.

2KCl + H2SO4 = 2KCl + 2HCl.

n (HCl) = n (KCl) = n (NaCl) = 4 mol.

M (NaCl) = 58.5 mol.

M (KCl) = 74.5 mol.

Let the amount of substance KCl – x mol; NaCl – y mol.

We get the expression:

x + y = 4 mol.

m = n × M.

m (KCl) = 74.5 x;

m (NaCl) = 58.5y.

74.5 x + 58.5y = 261.

Let’s compose a system of equations.

{x + y = 4

{74.5 x + 58.5 y = 261,

x = 4 – y

74.5 (4 – y) + 58.5y = 261.

298 – 74.5y + 58.5y = 261

298 – 261 = 74.5y – 58.5y

37 = 16y

y = 37: 16

y = 2.3

x = 4 – 2.3.

x = 1.7.

Let’s find the mass of KCl.

m = 1.7 × 74.5 = 126.65.

Find the mass of NaCl.

m = 2.3 × 58.5 = 134.55.

Let’s find the mass fractions of substances in the mixture.

w (KCl) = (126.65: 261) × 100% = 48.5%.

w (NaCl) = (134.55: 261) × 100% = 51.55%.

Answer: 51.55%; 48.5%.