The Haber procedure can be supplied to develop ammonia, #NH_3#, and also it is based upon the following reaction.

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

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

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The an initial thing the you must do below is to calculate the theoretical yield that the reaction, i.e. What you get if the reaction has actually a #100%# yield.

The balanced chemical equation

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

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

In your case, you understand that #1# mole of nitrogen gas reacts v #1# mole that hydrogen gas. Due to the fact that you don"t have sufficient hydrogen gas come ensure the all the moles that nitrogen gas can react

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

you deserve to say the hydrogen gas will certainly act together a limiting reagent, i.e. It will be completely consumed before every the moles of nitrogen gas will gain the opportunity to take part in the reaction.

So, the reaction will certainly consume #1# mole the 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 mole NH"_3#

at #100%# yield. This represents the reaction"s theoretical yield.

Now, you recognize that the reaction developed #0.50# moles of ammonia. This to represent the reaction"s actual yield.

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In bespeak to find the percent yield, you need to figure out how plenty of moles the ammonia room actually produced for every #100# moles of ammonia the could theoretically it is in produced.

You understand that #0.667# moles will create #0.50# moles, therefore you can say that

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