Chapter 4, Problem 124AP

A 0.8870-g sample of a mixture of $\text{NaCl}$ and $\text{KCl}$ is dissolved in water, and the solution is then treated with an excess of ${\text{AgNO}}_{\text{3}}$ to yield 1.913 g of $\text{AgCl}$. Calculate the percent by mass of each compound in the mixture.

Interpretation Introduction

Interpretation:

The percentage of $\text{NaCl}$ and $\text{KCl}$ in the given mixture is to be calculated.

Concept introduction:

Percentage of the given compound in a mixture is calculated by dividing the mass of the compound by the mass of the mixture and then multiplying it with $100$.

The number of moles of the compound formed in the given reaction is calculated by dividing its mass by the molar mass.

$\text{Moles of compound }=\frac{\text{mass of compund }}{\text{molar mass of compound}}$

Answer to Problem 124AP

Solution: $\text{NaCl}$ is $44.11\%$, and $\text{KCl}$ is $55.89\%$.

Explanation of Solution

Given information: The reaction of $\text{NaCl}$ and $\text{KCl}$ solution with ${\text{AgNO}}_3$ leads to the formation of $\text{AgCl}$.

The mass of the mixture is $0.8870\text{g}$.

The mass of $\text{AgCl}$ formed in the reaction is $1.913\text{g}$.

The balanced chemical equation for the given reaction is as follows:

$\text{NaCl}\left(aq\right)+\text{KCl}\left(aq\right)+{\text{AgNO}}_3\left(aq\right)\to \text{AgCl}\left(s\right)+{\text{NaNO}}_3\left(aq\right)+{\text{KNO}}_3\left(aq\right)$

The number of moles of $\text{AgCl}$ formed in the given reaction is calculated by dividing its mass by the molar mass.

$\begin{array}{c}\text{Moles}\text{of}\text{AgCl}=\frac{1.913\text{g}}{143.35\text{g/mol}}\\ =0.013345\text{mol}\end{array}$

The number of moles of $\text{AgCl}$ formed in the given reaction is equal to the sum of the number of moles of $\text{NaCl}$ and $\text{KCl}$ present in the given reaction.

The number of moles of $\text{NaCl}$ present in the given reaction is considered to be $x$. Hence, the number of moles of $\text{KCl}$ in the given reaction is $\left(0.013345\text{mol}-x\right)$.

The mass of the $\text{NaCl}$ and $\text{KCl}$ present in the given reaction is calculated by the formula,

$\text{Mass of compound=}\text{moles of compound}\times \text{Molar}\text{mass of compound}$ …… (1)

Substitute the values of number of moles and molar mass for $\text{NaCl}$ in the equation (1),

$\begin{array}{c}\text{Mass}\text{of}\text{NaCl}=x\text{mol}\times 58.44\text{g/mol}\\ =58.44x\text{g}\end{array}$ …… (2)

Substitute the values of number of moles and molar mass for $\text{KCl}$ in the equation (1),

$\begin{array}{c}\text{Mass}\text{of}\text{KCl}=\left(0.013345-x\right)\text{mol}\times 74.55\text{g/mol}\\ =74.55\left(0.013345-x\right)\text{g}\end{array}$ …... (3)

The number of moles of $\text{NaCl}$ present in the given reaction is calculated by the formula,

$\text{Mass}\text{of}\text{NaCl}+\text{Mass}\text{of}\text{KCl}=\text{Mass}\text{of}\text{AgCl}$

Substitute the values of mass of $\text{NaCl,}\text{KCl}$ and $\text{AgCl}$ in the above equation,

$\begin{array}{c}\left(58.44x\text{g}+74.55\left(0.013345-x\right)\text{g}\right)=0.8870\text{g}\\ \left(58.44x-74.55x\right)=\left(0.8870-0.9948\right)\\ x=\left(\frac{0.1078}{16.11}\right)\\ =6.6914\times {10}^{-3}\text{mol}\end{array}$

Hence, the number of moles of $\text{KCl}$ in the given reaction is calculated as,

$\begin{array}{c}\text{Moles}\text{of}\text{KCl}=0.013345\text{mol}-6.6958\times {10}^{-3}\text{mol}\\ =\text{6}\text{.6492}\times {10}^{-3}\text{mol}\end{array}$

Hence, the mass of $\text{NaCl}$ in the given reaction is calculated by substituting the value of $x$ in the equation (2),

$\begin{array}{c}\text{Mass}\text{of}\text{NaCl}=58.44\text{g/mol}\times 6.6914\times {10}^{-3}\text{mol}\\ =\text{0}\text{.3913}\text{g}\end{array}$

Similarly, the mass of $\text{KCl}$ in the given reaction is calculated by substituting the value of $\left(0.13345-x\right)$ in the equation (3),

$\begin{array}{c}\text{Mass}\text{of}\text{KCl}=74.55\text{g/mol}\times 6.6492\times {10}^{-3}\text{mol}\\ =\text{0}\text{.4957}\text{g}\end{array}$

The percentage of the compound in the given mixture is calculated by the formula,

$\text{Percentage}\text{of}\text{compound}=\frac{\text{Mass}\text{of}\text{compound}}{\text{Mass}\text{of}\text{mixture}}\times 100$ …… (4)

Substitute the values of mass of $\text{NaCl}$ and mass of mixture in the equation (4),

$\begin{array}{c}\text{Percentage}\text{of}\text{compound}=\frac{\text{0}\text{.3913}\text{g}}{\text{0}\text{.8870}\text{g}}\times 100\\ =44.11\%\end{array}$

Substitute the values of mass of $\text{KCl}$ and mass of mixture in the equation (4),

$\begin{array}{c}\text{Percentage}\text{of}\text{compound}=\frac{\text{0}\text{.4957}\text{g}}{\text{0}\text{.8870}\text{g}}\times 100\\ =55.89\%\end{array}$

Conclusion

The percentage by mass of $\text{NaCl}$ and $\text{KCl}$ in a 0.8870 gm sample mixture of both compound is $44.11\%$ and $55.89\%$, respectively.