# A reaction of 53.1 "g" of "Na" and 35.7 "g" of "Br"_2 yields 40 "g" of "NaBr". What is the percentage yield?

Nov 17, 2016

16.8%

#### Explanation:

In order to work out the percentage yield of a reaction, we must use the following formula:

% "yield": "yield from reaction"/("theoretical yield")xx100

Now we must work out the theoretical yield of $\text{NaBr}$

We start by writing a balanced equation for the reaction

$2 \text{Na" color(white)(a)+ color(white)(a) "Br"_2 color(white)(a)rarr color(white)(a)2"NaBr}$

As we can see, the molar ratio of $\text{Na " : " NaBr}$ is $1 : 1$, so we shall use these two molecules to work out the theoretical yield.

Now that we have the balanced equation, we must work out how many moles we have of any reagent (in this case, sodium).

Moles can be worked out using this formula:

"n" ("moles")=("m " ("mass"))/("M (atomic mass)")

For sodium:

$\text{n} = \frac{53.1}{23} = 2.31$

Now that we have the moles of sodium, we can work out the moles of sodium bromide. Since the molar ratio is $1 : 1$, the rules of stoichiometry tell us that we also have $2.31$ moles of $\text{NaBr}$.

We have the moles and the atomic mass of $\text{NaBr}$, but we don't have the mass. However, we can easily rearrange the formula for moles.

$\text{n"="m"/"M}$

$\text{m"="n"xx"M"=2.31xx103=237.93 " g}$

Finally, we can work out the percentage yield

% " yield"=40/237.93xx100=16.8%