Question

Question #c4ab3

Answer 1

`"% 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.