What is the percent yield for the following chemical reaction?

Question

The Haber process can be used to produce ammonia, #NH_3#, and it is based on the following reaction.

#N_2(g)+3H_2(g)->2NH_3(g)#

If one mole each of #N_2# and #H_2# are mixed and 0.50 moles of #NH_3# are produced, what is the percent yield for the reaction?

Wyatt Atkins · Accepted Answer

#75%#

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 chemical equation in balance

#"N"_ (2(g)) + 3"H"_ (2(g)) -> 2"NH"_ (3(g))#

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"))#

It is possible to describe hydrogen gas as a limiting reagent, meaning that it will be used up entirely before all the moles of nitrogen gas have an opportunity to participate 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"#

As a result, the reaction's percent yield can be expressed as

#color(darkgreen)(ul(color(black)("% yield = 75%")))#

I'll round the result to two significant figures.

Wyatt Adkins · Answer

undefined To calculate the percent yield for a chemical reaction, you need to divide the actual yield (the amount of product obtained in the experiment) by the theoretical yield (the maximum amount of product that could be obtained based on stoichiometry), and then multiply by 100 to express it as a percentage. The formula is:

Percent Yield = (Actual Yield / Theoretical Yield) * 100

What is the percent yield for the following chemical reaction?

The Haber process can be used to produce ammonia, #NH_3#, and it is based on the following reaction. #N_2(g)+3H_2(g)->2NH_3(g)# If one mole each of #N_2# and #H_2# are mixed and 0.50 moles of #NH_3# are produced, what is the percent yield for the reaction?

The Haber process can be used to produce ammonia, #NH_3#, and it is based on the following reaction.

#N_2(g)+3H_2(g)->2NH_3(g)#

If one mole each of #N_2# and #H_2# are mixed and 0.50 moles of #NH_3# are produced, what is the percent yield for the reaction?