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Chapter 4: Problem 140

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

Short Answer

Expert verified
The percent by mass of \(NaCl\) in the mixture is 65.206 \% and the percent by mass of \(KCl\) in the mixture is 34.794 \%

Step by step solution

01

Calculate the moles of \(AgCl\)

The molar mass of \(AgCl\) is \(143.32 \, g/mol\). Thus, the number of moles of \(AgCl\) produced is \(1.913 \, g / 143.32 \, g/mol = 0.01336 \, mol\)
02

Determine the moles of \(Cl^-\) in the mixture

In the reactions \(NaCl + AgNO_3 \rightarrow AgCl + NaNO_3\) and \(KCl + AgNO_3 \rightarrow AgCl + KNO_3\), one mole of \(Cl^-\) gives one mole of \(AgCl\). Therefore, the moles of \(Cl^-\) in the mixture are also 0.01336 mol.
03

Calculate the mass of the \(Cl^-\) in the mixture

The molar mass of \(Cl^-\) is \(35.45 \, g/mol\). Thus, its mass in the mixture is \(0.01336 \, mol \times 35.45 \, g/mol = 0.4733 \, g\)
04

Determine the mass of the \(NaCl\) and \(KCl\) in the mixture

The mass of the mixture is the sum of the masses of \(NaCl\) and \(KCl\). Thus, the mass of \(NaCl\) and \(KCl\) is \(0.8870 \, g + 0.4733 \, g = 1.3603 \, g\)
05

Calculate the percent by mass of each compound in the mixture

The percent by mass of \(NaCl\) is \((0.8870 \, g / 1.3603 \, g) \times 100 = 65.206 \%\) and the percent by mass of \(KCl\) is \((0.4733 \, g / 1.3603 \, g) \times 100 = 34.794 \%\)

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Most popular questions from this chapter

How would you prepare \(60.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HNO}_{3}\) from a stock solution of \(4.00 \mathrm{M} \mathrm{HNO}_{3}\) ?

A \(1.00-\mathrm{g}\) sample of a metal \(\mathrm{X}\) (that is known to form \(\mathrm{X}^{2+}\) ions ) was added to \(0.100 \mathrm{~L}\) of \(0.500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\). After all the metal had reacted, the remaining acid required \(0.0334 \mathrm{~L}\) of \(0.500 \mathrm{M} \mathrm{NaOH}\) solution for neutralization. Calculate the molar mass of the metal and identify the element.

A \(5.00 \times 10^{2} \mathrm{~mL}\) sample of \(2.00 \mathrm{M} \mathrm{HCl}\) solution is treated with \(4.47 \mathrm{~g}\) of magnesium. Calculate the concentration of the acid solution after all the metal has reacted. Assume that the volume remains un- changed.

Ammonium nitrate \(\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)\) is one of the most important nitrogen-containing fertilizers. Its purity can be analyzed by titrating a solution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) with a standard \(\mathrm{NaOH}\) solution. In one experiment a \(0.2041-\mathrm{g}\) sample of industrially prepared \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) required \(24.42 \mathrm{~mL}\) of \(0.1023 \mathrm{M} \mathrm{NaOH}\) for neutralization. (a) Write a net ionic equation for the reaction. (b) What is the percent purity of the sample?

A \(46.2-\mathrm{mL}, 0.568 M\) calcium nitrate \(\left[\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\right]\) solution is mixed with \(80.5 \mathrm{~mL}\) of \(1.396 \mathrm{M}\) calcium nitrate solution. Calculate the concentration of the final solution.
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