# In a double replacement reaction between silver nitrate and magnesium bromide. Calculate the mass of silver bromide which should be produced from 22.5g of silver nitrate. The actual yield was 22.0g, how do you calculate percent yield?

##### 1 Answer
Oct 29, 2015

Theoretical yield: $\text{24.9 g}$
Percent yield: $\text{88.4%}$

#### Explanation:

Start by writing the balanced chemical equation for this double replacement reaction

2"AgNO"_text(3(aq]) + "MgBr"_text(2(aq]) -> color(red)(2)"AgBr"_text((s]) darr + "Mg"("NO"_3)_text(2(aq])

Notice that you have a $1 : \textcolor{red}{2}$ mole ratio between silver nitrate and silver bromide. This means that the reaction will produce $\textcolor{red}{1}$ mole of the latter for every $1$ moel of the former that takes part in the reaction.

However, keep in mind that this is true for a reaction that has a 100% yield - this represents the theoretical yield of the reaction.

If the percent yield of the reaction is smaller than 100%, than silver bromide will not be produced in a $1 : \textcolor{red}{1}$ mole ratio with silver nitrate.

So, how many moles of silver nitrate do you have in $\text{22,5 g}$ of silver nitrate? Use the compound's molar mass to get

22.5color(red)(cancel(color(black)("g"))) * "1 mole AgNO"_3/(169.87color(red)(cancel(color(black)("g")))) = "0.13245 moles AgNO"_3

So, what would the theoretical yield of the reaction be?

Well, the $1 : 1$ mole ratio would produce

0.13245color(red)(cancel(color(black)("moles AgNO"_3))) * "1 mole AgBr"/(1color(red)(cancel(color(black)("mole AgNO"_3)))) = "0.13245 moles AgBr"

To calculate how many grams of silver bromide would contain this many moles? Once again, use the compound's molar mass

0.13245color(red)(cancel(color(black)("moles"))) * "187.77 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("24.9 g AgBr")

Now, a reaction's percent yield is defined as

color(blue)(%"yield" = "actual yield"/"theoretical yield" xx 100)

If the actual yield is $\text{22.0 g}$, then the percent yield of the reaction is

"% yield" = (22.0color(red)(cancel(color(black)("g"))))/(24.9color(red)(cancel(color(black)("g")))) xx 100 = color(green)("88.4%")

Here's a photo of how the silver bromide precipitate would look like