# "1.95 g" of "H"_2 is allowed to react with "9.94 g" of "N"_2, producing "1.56 g" of "NH"_3. What is the theoretical yield in grams for this reaction under the given conditions?

## What is the percent yield for this reaction under the given conditions?

Aug 24, 2017

Here's what I got.

#### Explanation:

Start by writing the balanced chemical equation that describes this reaction

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

Notice that for every $1$ mole of nitrogen gas that takes part in the reaction, the reaction consumes $3$ moles of hydrogen gas and produces $2$ moles of ammonia.

Now, the first thing that you need to do here is to figure out if you're dealing with a limiting reagent.

Use the molar masses of the two reactants to convert the masses to moles

1.95 color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.016color(red)(cancel(color(black)("g")))) = "0.9673 moles H"_2

9.94 color(red)(cancel(color(black)("g"))) * "1 mole N"_2/(28.0134color(red)(cancel(color(black)("g")))) = "0.3548 moles N"_2

Now, in order for all the moles of nitrogen gas to take part in the reaction, you would need

0.3548 color(red)(cancel(color(black)("moles N"_2))) * "3 moles H"_2/(1color(red)(cancel(color(black)("mole N"_2)))) = "1.0644 moles H"_2

In your case, you don't have enough moles of hydrogen gas to allow for this to happen

overbrace("1.0644 moles H"_2)^(color(blue)("what you need")) " " > " " overbrace("0.9673 moles H"_2)^(color(blue)("what you have"))

so you can say that hydrogen gas will act as the limiting reagent, i.e. it will be completely consumed by the reaction before all the moles of nitrogen gas will have the chance to react.

You can thus say that the reaction will consume $0.9673$ moles of hydrogen gas and produce

0.9673 color(red)(cancel(color(black)("moles H"_2))) * "2 moles NH"_3/(3color(red)(cancel(color(black)("moles H"_2)))) = "0.6449 moles NH"_3

To convert this to grams, use the molar mass of ammonia

$0.6449 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles NH"_3))) * "17.031 g"/(1color(red)(cancel(color(black)("mole NH"_3)))) = color(darkgreen)(ul(color(black)("11.0 g}}}}$

The answer is rounded to three sig figs.

So, this represents the theoretical yield of the reaction, i.e. what you would expect the reaction to produce at 100% yield.

In your case, you know that the reaction produced $\text{1.56 g}$ of ammonia, so your goal now is to figure out the number of grams of ammonia that the reaction produces for every $\text{100 g}$ of ammonia that it could theoretically produce, i.e. the percent yield of the reaction.

100 color(red)(cancel(color(black)("g NH"_3color(white)(.)"in theory"))) * ("1.56 g NH"_3color(white)(.)"produced")/(11.0color(red)(cancel(color(black)("g NH"_3color(white)(.)"in theory")))) = "14.2 g NH"_3color(white)(.)"produced"

You can thus say that the percent yield of the reaction is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{% yield = 14.2%}}}}$

Once again, the answer is rounded to three sig figs.