Question

When 10g of a mixture of potassium chloride and anhydrous sodium sulphate is dissolved in water and excess barium chloride solution added, 6.9g of barium sulphate is precipitated. Calculate the composition of the mixture

Answer 1

Upon addition of the barium chloride to a mixture of potassium chloride and sodium sulphate, only sodium sulphate reacts with barium chloride. The balanced equation of the reaction between sodium sulphate and barium chloride is:

Na₂SO₄ + BaCl₂ $\rightarrow$ 2NaCl + BaSO₄$\downarrow$ .

According to the stoichiometric coefficients, the number of the moles of sodium sulphate that react and the number of the moles of barium sulphate that is formed relate as:

$n(Na_2SO_4) = n(BaSO_4)$.

The number of the moles of barium sulphate can be calculated from its mass using its molar weight $M(BaSO_4) = 233.39$ g/mol:

$n(BaSO_4) = \frac{m}{M} = \frac{6.9 \text{ g}}{233.39 \text{ g/mol}} =0.02956$ mol.

Therefore, the number of the moles of sodium sulphate and its mass ($M(Na_2SO_4) = 142.04$ g/mol) are:

$n(Na_2SO_4) = 0.02956$ mol,

$m(Na_2SO_4) = n·M$

$m(Na_2SO_4)= 0.02956\text{ mol}·142.04\text{ g/mol} = 4.2$ g.

Therefore, the composition of the mixture is the following: 4.2 g of Na₂SO₄ and 10-4.2 = 5.8 g of KCl.

Answer: 4.2 g of Na₂SO₄ and 5.8 g of KCl.