Question

A 0.9140 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO₃ to yield 1.943 g of AgCl. Calculate the percent by mass of each compound in the mixture.

Answer 1

Let the sample contain x g NaCl and y g KCl; therefore,

x + y = 0.9140 …….(1).

Write down the balanced chemical equations for the reactions.

NaCl (aq) + AgNO₃ (aq) --------> AgCl (s) + NaNO₃ (aq).

KCl (aq) + AgNO₃ (aq) -------> AgCl (s) + KNO₃ (aq).

As per the stoichiometric equations,

1 mole AgCl = 1 mole NaCl = 1 mole KCl.

We shall need the molar masses of NaCl (58.44 g/mol), KCl (74.5513 g/mol) and AgCl (143.32 g/mol).

Therefore, x g NaCl = ( x g/58.44 g/mol) = (x/58.44) mole NaCl = (x/58.44) mole AgCl.

y g KCl = (y g/74.5513 g/mol) = (y/74.5513) mole KCl = (y/74.5513) mole AgCl.

Total mole(s) of AgCl = (x/58.44 + y/74.5513) mole.

Total mass of AgCl = (143.32 g/mol)*(x/58.44 + y/74.5513) mole = (143.32)*(x/58.44 + y/74.6613) g.

Therefore,

(143.32)*(x/58.44 + y/74.5513) = 1.943

====> x/58.44 + y/74.5513 = 0.013557

====> x/58.44 + (0.9140 – x)/(74.5513) = 0.013557

====> x/58.44 + 0.9140/74.5513 – x/74.5513 = 0.013557

====> x/58.44 + 0.012260 – x/74.5513 = 0.013557

====> x/58.44 – x/74.5513 = 0.001297

====> 74.5513x – 58.44x = 0.001297*58.44*74.5513 = 5.65074

====> 16.1113x = 5.65074

====> x = 0.3507

Percent NaCl in the mixture = (0.3507 g)/(0.9140 g)*100 = 38.3698% ≈ 36.37% (ans).

Percent KCl in the mixture = (0.9140 – 0.3507)g/(0.9140 g)*100 = 61.6302% ≈ 61.63% (ans).