#"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?
What is the percent yield for this reaction under the given conditions?
1 Answer
Here's what I got.
Explanation:
Start by writing the balanced chemical equation that describes this reaction
#"N"_ (2(g)) + 3"H"_ (2(g)) -> 2"NH"_ (3(g))#
Notice that for every
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 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 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")))#
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
In your case, you know that the reaction produced
#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
#color(darkgreen)(ul(color(black)("% yield = 14.2%")))#
Once again, the answer is rounded to three sig figs.