#"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?

Answer 1

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.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email

Answer 2

Henry Antrim

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

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.