# Question c4ab3

Sep 6, 2017

"% yield" = 68.5826%

#### Explanation:

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

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

The reaction consumes sodium chloride and silver nitrate in a $1 : 1$ mole ratio and produces silver chloride in $1 : 1$ mole ratios to the two reactants, so you can say that at 100% yield, you have

overbrace("moles of NaCl = moles of AgNO"_3)^(color(blue)("consumed by the reaction")) = overbrace("moles of AgCl")^(color(blue)("produced by the reaction"))

Use the molar mass of sodium chloride to convert the mass to moles

53.7648 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))) = "0.9200 moles NaCl"

Since silver nitrate is in excess, you can say that at 100% yield, the reaction will consume $0.9200$ moles of sodium chloride and produce $0.9200$ moles of silver chloride.

To convert the number of moles to grams, use the molar mass of silver chloride

0.9200 color(red)(cancel(color(black)("moles AgCl"))) * "143.32 g"/(1color(red)(cancel(color(black)("mole AgCl")))) = "131.8544 g"

However, you know that the reaction produced $\text{90.42919 g}$ of silver chloride, less than what you would expect to see at 100% yield.

To find the percent yield of the reaction, calculate the mass of silver chloride produced for every $\text{100 g}$ of silver chloride that could theoretically be produced by the reaction

100 color(red)(cancel(color(black)("g in theory"))) * "90.42919 g produced"/(131.8544color(red)(cancel(color(black)("g in theory")))) = "68.5826 g"

Therefore, you can say that the reaction will have a percent yield of

color(darkgreen)(ul(color(black)("% yield" = 68.5826%)))#

The answer is rounded to six sig figs, the number of sig figs you have for the mass of sodium chloride.