#"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?

Question

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

Mateo Aguilar · Accepted Answer

Here's what I got.

Write the balanced chemical equation describing this reaction first. #"N"_ (2(g)) + 3"H"_ (2(g)) -> 2"NH"_ (3(g))# 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. Determining whether you're working with a limiting reagent is the first thing you need to do in this situation. To convert the masses to moles, utilize the molar masses of the two reactants. #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# Thus, for every mole of nitrogen gas to participate in the reaction, you would require #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# You don't have enough moles of hydrogen gas in your situation for this to occur. #overbrace("1.0644 moles H"_2)^(color(blue)("what you need")) " " > " " overbrace("0.9673 moles H"_2)^(color(blue)("what you have"))# Thus, it can be said that hydrogen gas will function as the limiting reagent because the reaction will use up all of it before any moles of nitrogen gas have a 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# This can be converted to grams using the ammonia molar mass. #0.6449 color(red)(cancel(color(black)("moles NH"_3))) * "17.031 g"/(1color(red)(cancel(color(black)("mole NH"_3)))) = color(darkgreen)(ul(color(black)("11.0 g")))# Three sig figs are used to round the result. 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 #"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 #"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"# Thus, the reaction's percent yield can be expressed as follows: #color(darkgreen)(ul(color(black)("% yield = 14.2%")))# The result is rounded to three significance figures once more.

Henry Antrim · Answer

undefined The theoretical yield for this reaction is 7.47 grams of NH3.