Chapter 4, Problem 4.140QP

A 0.8870-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 1.913 g of AgCl. Calculate the percent by mass of each compound in the mixture.

Interpretation Introduction

Interpretation:

The mass percent of $\text{NaCl}$ and $\text{KCl}$ in given $\text{NaCl}$ and $\text{KCl}$ mixture should be calculated.

Concept introduction:

Precipitation reaction:

If precipitate is formed, when two soluble salt solutions are mixed together is known as precipitation reaction.

The mass of the precipitate is depends on the reactant masses.

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

The mole of compound is given by the ratio between taken mass of the compound and molar mass of the compound.

$\text{Mole}\text{=}\frac{\text{Massof}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}$

Mass percentage:

Mass percentage of given compound is calculated by the ration between the mass of the analyte and total mass of the compound.

$\text{Mass}\text{percent}\text{=}\frac{\text{Mass}}{\text{Total}\text{mass}}$

Answer to Problem 4.140QP

• The mass percent of $\text{NaCl}$ in given $\text{0}\text{.8870}\text{g}$ mixture is $\text{44}\text{.11}\text{\%}$
• The mass percent of $\text{KCl}$ in given $\text{0}\text{.8870}\text{g}$ mixture is $\text{55}\text{.89}\text{\%}$

Explanation of Solution

Record the given data,

Mass of $\text{NaCl}$ and $\text{KCl}$ mixture = $\text{0}\text{.8870}\text{g}$

Mass of produced $\text{AgCl}$ precipitate = $\text{1}\text{.913 g}$

The given masses are recorded as shown above.

The mole of $\text{KCl}$ and $\text{NaCl}$ mixture to produce a $\text{1}\text{.913 g}$$\text{AgCl}$ precipitate

Molar mass of $\text{AgCl}$ is $\text{143}\text{.35}\text{g}$

The balance equation is,

$\begin{array}{l}{\text{XCl(aq)+AgNO}}_{\text{3(aq)}}\to {\text{AgCl}}_{\text{(s)}}{\text{+XNO}}_{\text{3}}{}_{\text{(aq)}}\\ \text{Where}\text{X=}\text{Na}\text{or}\text{K}\end{array}$

$\begin{array}{l}\text{XCl}\text{mole}\text{=}\text{1}\text{.913}\text{g}\text{AgCl}\text{×}\frac{\text{1}\text{mol}\text{AgCl}}{\text{143}\text{.35}\text{g}}\text{×}\frac{\text{1}\text{mol}\text{XCl}}{\text{1}\text{mol AgCl}}\\ \text{=}\text{0}\text{.013345}\text{mol}\text{XCl}\end{array}$

• One mole of $\text{XCl}$ is required to react with one mole of ${\text{AgNO}}_{\text{3}}$ to produce a mole of $\text{AgCl}$ precipitate.
• The mass of $\text{AgCl}$ and molar mass of $\text{AgCl}$ are plugged in the above equation to give the mole of  $\text{KCl}$ and $\text{NaCl}$ mixture to produce a $\text{1}\text{.913 g}$$\text{AgCl}$ precipitate
• The mole of $\text{KCl}$ and $\text{NaCl}$ mixture to produce a $\text{1}\text{.913 g}$$\text{AgCl}$ precipitate is $\text{0}\text{.013345}\text{mol}\text{XCl}\text{.}$

The moles of $\text{KCl}$ and  $\text{NaCl}$ in give $\text{0}\text{.8870}\text{g}$ mixture

Let consider x=number of moles of $\text{NaCl}$ in give mixture, then the number of moles of $\text{KCl}$ is

$\text{KCl}\text{=}\text{0}\text{.013345}\text{mol-x}$

The sum of $\text{KCl}$ and $\text{NaCl}$ masses are equal to $\text{0}\text{.8870}\text{g}$

Molar mass of $\text{KCl}$ is $\text{74}\text{.55}\text{g}$

Molar mass of $\text{NaCl}$ is $\text{58}\text{.44}\text{g}$

$\begin{array}{l}\text{0}\text{.8870}\text{g}\text{=}\left[\text{x}\text{mol}\text{NaCl×}\frac{\text{58}\text{.44}\text{g}\text{NaCl}}{\text{1}\text{}\text{mole}\text{NaCl}}\right]\text{+}\left[\left(\text{0}\text{.013345-x}\right)\text{mol}\text{KCl×}\frac{\text{74}\text{.55}\text{g}\text{KCl}}{\text{1}\text{}\text{mol}\text{}\text{KCl}}\right]\\ \text{x}\text{=}\text{6}\text{.6958}{\text{×10}}^{\text{-3}}\text{moles}\text{NaCl}\\ \text{Mole}\text{KCl}\text{=}\text{0}\text{.013345-}\text{x}\\ \text{=}\text{0}\text{.013345-6}\text{.6958}{\text{×10}}^{\text{-3}}\\ \text{=}\text{6}\text{.6492}{\text{×10}}^{\text{-3}}\text{mol}\text{KCl}\end{array}$

• The calculated mole of $\text{KCl}$ and $\text{NaCl}$ mixture and molar masses of $\text{KCl}$ and $\text{NaCl}$ are plugged in above equation and do some maths to give the mole of $\text{KCl}$ and $\text{NaCl}$ present in give $\text{0}\text{.8870}\text{g}$ mixture.
• The mole of $\text{KCl}$ present in give $\text{0}\text{.8870}\text{g}$ mixture is $\text{6}\text{.6492}{\text{×10}}^{\text{-3}}$
• The mole of $\text{NaCl}$ present in give $\text{0}\text{.8870}\text{g}$ mixture is $\text{6}\text{.6958}{\text{×10}}^{\text{-3}}$

The masses of $\text{KCl}$ and  $\text{NaCl}$ in give $\text{0}\text{.8870}\text{g}$ mixture

$\begin{array}{l}\text{Mole}\text{of}\text{NaCl}\text{=}\text{(6}{\text{.6958×10}}^{\text{-3}}\text{mol}\text{NaCl)×}\frac{\text{58}\text{.44}\text{g}\text{NaCl}}{\text{1}\text{mol}\text{NaCl}}\\ \text{=}\text{0}\text{.3913}\text{g}\text{NaCl}\\ \\ \text{Mole}\text{of}\text{KCl}\text{=}\text{(6}{\text{.6492×10}}^{\text{-3}}\text{mol}\text{KCl)×}\frac{\text{74}\text{.55}\text{g}\text{KCl}}{\text{1}\text{mol}\text{KCl}}\\ \text{=}\text{0}\text{.4957}\text{g}\text{KCl}\end{array}$

• The calculated mole of $\text{KCl}$ and  $\text{NaCl}$ are multiplied with molar masses of $\text{KCl}$ and  $\text{NaCl}$ to give the masses of $\text{KCl}$ and  $\text{NaCl}$ in give $\text{0}\text{.8870}\text{g}$ mixture
• Mass of $\text{KCl}$ in give $\text{0}\text{.8870}\text{g}$ mixture is $\text{0}\text{.4957}\text{g}$
• Mass of $\text{NaCl}$ in give $\text{0}\text{.8870}\text{g}$ mixture is $\text{0}\text{.3913}\text{g}$

The mass percentages of $\text{KCl}$ and  $\text{NaCl}$ in give $\text{0}\text{.8870}\text{g}$ mixture

$\begin{array}{l}\text{Mass}\text{\%}\text{NaCl}\text{=}\frac{\text{0}\text{.3913}\text{g}}{\text{0}\text{.8870}\text{g}}\text{×100}\\ \text{=}\text{44}\text{.11}\text{\%}\\ \\ \text{Mass}\text{\%}\text{KCl}\text{=}\frac{\text{0}\text{.49}\text{.57}\text{g}}{\text{0}\text{.8870}\text{g}}\text{×100}\\ \text{=}\text{55}\text{.89}\text{\%}\end{array}$

• The calculated masses of $\text{KCl}$ and  $\text{NaCl}$ are divided by total mass of mixture $\text{0}\text{.8870}\text{g}$ to give a mass percentages of $\text{KCl}$ and  $\text{NaCl}$ in given mixture.
• The mass percent of $\text{NaCl}$ in given $\text{0}\text{.8870}\text{g}$ mixture is $\text{44}\text{.11}\text{\%}$
• The mass percent of $\text{KCl}$ in given $\text{0}\text{.8870}\text{g}$ mixture is $\text{55}\text{.89}\text{\%}$

Conclusion

The mass percent of $\text{NaCl}$ and $\text{KCl}$ in given $\text{NaCl}$ and $\text{KCl}$ mixture calculated.