Question

3.6g mixture of sodium chloride and potassium chloride is dissolved in water. The solution is treated with excess of silver nitrate solution. $7.74{\text{ g}}$ of silver chloride is obtained. Find the percentage of sodium chloride and potassium chloride in the mixture.

Answer 1

Hint: We have to begin with writing the balanced chemical equations for the reaction of both sodium chloride and potassium chloride with silver nitrate. Wrong balanced chemical equations give wrong answers.

Complete step by step solution:
The balanced chemical equation for the reaction of sodium chloride and silver nitrate is,
${\text{NaCl}} + {\text{AgN}}{{\text{O}}_3} \to {\text{AgCl}} + {\text{NaN}}{{\text{O}}_3}$
From the reaction stoichiometry, we can say that
$1{\text{ mol NaCl}} = 1{\text{ mol AgCl}}$
The mass of one mole of every substance is equal to its molar mass. Thus,
$58.44{\text{ g NaCl}} = 143.32{\text{ g AgCl}}$
Let the mass of ${\text{NaCl}}$ in the mixture be $x{\text{ g}}$. Thus, $x{\text{ g}}$ of ${\text{NaCl}}$ will produce,
${\text{Mass of AgCl}} = \dfrac{{143.32{\text{ g AgCl}}}}{{58.44{\text{ g NaCl}}}} \times x{\text{ g NaCl}}$ …… (1)
The balanced chemical equation for the reaction of potassium chloride and silver nitrate is,
${\text{KCl}} + {\text{AgN}}{{\text{O}}_3} \to {\text{AgCl}} + {\text{KN}}{{\text{O}}_3}$
From the reaction stoichiometry, we can say that
$1{\text{ mol KCl}} = 1{\text{ mol AgCl}}$
The mass of one mole of every substance is equal to its molar mass. Thus,
$74.55{\text{ g KCl}} = 143.32{\text{ g AgCl}}$
The mass of ${\text{KCl}}$ present in the mixture is,${\text{Mass of KCl in the mixture}} = {\text{Mass of mixture}} - {\text{Mass of NaCl in the mixture}}$
${\text{Mass of KCl in the mixture}} = \left( {{\text{3}}{\text{.6}} - x} \right){\text{g}}$
Thus, $\left( {{\text{3}}{\text{.6}} - x} \right){\text{g}}$ of ${\text{KCl}}$ will produce,
${\text{Mass of AgCl}} = \dfrac{{143.32{\text{ g AgCl}}}}{{74.55{\text{ g KCl}}}} \times \left( {{\text{3}}{\text{.6}} - x} \right){\text{g KCl}}$ ……(2)
Now, the mass of silver chloride obtained is equation (1) + equation (2). Thus,
$\dfrac{{143.32{\text{ g AgCl}}}}{{58.44{\text{ g NaCl}}}} \times x{\text{ g NaCl}} + \dfrac{{143.32{\text{ g AgCl}}}}{{74.55{\text{ g KCl}}}} \times \left( {{\text{3}}{\text{.6}} - x} \right){\text{g KCl}} = 7.74{\text{ g AgCl}}$
$2.450x + 1.921\left( {3.6 - x} \right) = 7.74$
$\Rightarrow 2.450x + 6.9156 - 1.921x = 7.74$
$\Rightarrow 2.450x - 1.921x = 7.74 - 6.951$
$\Rightarrow 0.529x = 0.825$
$\Rightarrow x = 1.56{\text{ g}}$
Thus, the mass of ${\text{NaCl}}$ in the mixture $ = x{\text{ g}} = {\text{1}}{\text{.56 g}}$.
Thus,
${\text{Mass of KCl in the mixture}} = \left( {{\text{3}}{\text{.6}} - x} \right){\text{g}}$
$\Rightarrow {\text{Mass of KCl in the mixture}} = \left( {{\text{3}}{\text{.6}} - 1.56} \right){\text{g}}$
$\Rightarrow {\text{Mass of KCl in the mixture}} = 2.04{\text{ g}}$
Percentage of ${\text{NaCl}}$ in the mixture $ = \dfrac{{1.56}}{{3.6}} \times 100\% $
$\Rightarrow$ Percentage of ${\text{NaCl}}$ in the mixture $ = 43.3\% $
$\Rightarrow$ Percentage of ${\text{KCl}}$ in the mixture $ = \dfrac{{2.04}}{{3.6}} \times 100\% $
$\Rightarrow$ Percentage of ${\text{KCl}}$ in the mixture $ = 56.7\% $

Thus, the percentage of sodium chloride in the mixture is $43.3\% $ and the percentage of potassium chloride in the mixture is $56.7\% $.

Note: The percentage of sodium chloride and potassium chloride in the mixture can be calculated by dividing the respective masses by total mass. The sum of the percentages should be 100. Please be careful while finding out the number of moles and then the percentage of each.