# What is the percent yield for the following chemical reaction?

## The Haber process can be used to produce ammonia, $N {H}_{3}$, and it is based on the following reaction. ${N}_{2} \left(g\right) + 3 {H}_{2} \left(g\right) \to 2 N {H}_{3} \left(g\right)$ If one mole each of ${N}_{2}$ and ${H}_{2}$ are mixed and 0.50 moles of $N {H}_{3}$ are produced, what is the percent yield for the reaction?

Sep 19, 2017

75%

#### Explanation:

The first thing that you need to do here is to calculate the theoretical yield of the reaction, i.e. what you get if the reaction has a 100% yield.

The balanced chemical equation

${\text{N"_ (2(g)) + 3"H"_ (2(g)) -> 2"NH}}_{3 \left(g\right)}$

tells you that every $1$ mole of nitrogen gas that takes part in the reaction will consume $3$ moles of hydrogen gas and produce $1$ mole of ammonia.

In your case, you know that $1$ mole of nitrogen gas reacts with $1$ mole of hydrogen gas. Since you don't have enough hydrogen gas to ensure that all the moles of nitrogen gas can react

overbrace("3 moles H"_2)^(color(blue)("what you need")) " " > " " overbrace("1 mole H"_2)^(color(blue)("what you have"))

you can say that hydrogen gas will act as a limiting reagent, i.e. it will be completely consumed before all the moles of nitrogen gas will get the chance to take part in the reaction.

So, the reaction will consume $1$ mole of hydrogen gas and produce

1 color(red)(cancel(color(black)("mole H"_2))) * "2 moles NH"_3/(3color(red)(cancel(color(black)("moles H"_2)))) = "0.667 moles NH"_3

at 100% yield. This represents the reaction's theoretical yield.

Now, you know that the reaction produced $0.50$ moles of ammonia. This represents the reaction's actual yield.

In order to find the percent yield, you need to figure out how many moles of ammonia are actually produced for every $100$ moles of ammonia that could theoretically be produced.

You know that $0.667$ moles will produce $0.50$ moles, so you can say that

100 color(red)(cancel(color(black)("moles NH"_3color(white)(.)"in theory"))) * ("0.50 moles NH"_3color(white)(.)"actual")/(0.667color(red)(cancel(color(black)("moles NH"_3color(white)(.)"in theory")))) = "75 moles NH"_3color(white)(.)"actual"

Therefore, you can say that the reaction has a percent yield equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{% yield = 75%}}}}$

I'll leave the answer rounded to two sig figs.