Question #c4ab3

1 Answer
Sep 6, 2017

Answer:

#"% yield" = 68.5826%#

Explanation:

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

#"AgNO"_ (3(aq)) + "NaCl"_ ((aq)) -> "AgCl"_ ((s)) darr + "NaNO"_ (3(aq))#

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 #"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 #"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.