# Question 0ca2b

Oct 16, 2016

Here's what I got.

#### Explanation:

Start by writing the balanced chemical equation that describes this double replacement reaction

${\text{KCl"_ ((aq)) + "AgNO"_ (3(aq)) -> "AgCl"_ ((s)) darr + "KNO}}_{3 \left(a q\right)}$

As you can see, every mole of potassium chloride that takes part in the reaction consumes $1$ mole of silver nitrate and produces $1$ mole of silver chloride, an insoluble solid that precipitates out of solution.

This means that you can use the number of moles of silver chloride produced by the reaction to find the number of moles of potassium chloride present in the original sample.

To do that, convert the grams of silver chloride produced to moles by using the compound's molar mass

0.8402 color(red)(cancel(color(black)("g"))) * "1 mole AgCl"/(143.32 color(red)(cancel(color(black)("g")))) = "0.0058624 moles AgCl"

This means that the original sample contained

0.0058624 color(red)(cancel(color(black)("moles AgCl"))) * "1 mole KCl"/(1color(red)(cancel(color(black)("mole AgCl")))) = "0.0058624 moles KCl"

To convert the number of moles of potassium chloride to grams, use the compound's molar mass

0.0058624color(red)(cancel(color(black)("moles KCl"))) * "74.55 g"/(1color(red)(cancel(color(black)("mole KCl")))) = "0.43704 g"

To find the percent of potassium chloride in the original sample, simply do

"% KCl" = (0.43704 color(red)(cancel(color(black)("g"))))/(0.4500color(red)(cancel(color(black)("g")))) xx 100% = color(green)(bar(ul(|color(white)(a/a)color(black)(97.12%)color(white)(a/a)|)))#

The answer is rounded to four sig figs.