Nitric oxide, #"NO"#, is made from the oxidation of #"NH"_3# as follows: #4"NH"_3 + 5"O"_2-> 4"NO"+ 6 "H"_2"O"#. If #"9.6 g"# of #"NH"_3# gives #"12.0 g"# of #"NO"#, what is the percent yield of #"NO"#?

1 Answer
Truong-Son N.
Jan 24, 2016

You should assume that oxygen is in excess because you are told ahead of time that you yield a specific amount of product.

What you have been given for the masses are numbers that you can use to get the actual yield for #"NO"# based on the amount of #"NH"_3#---the limiting reagent---that you have.

#color(green)("Actual Yield")# #=# #color(green)("12.0 g NO")#

Based on the reaction stoichiometry, let's see how much of #"NO"# product you SHOULD have made.

The mass of product #"NO"# can be calculated by doing the following conversions in sequence:

  1. From #"g"# of #"NH"_3# to #"mol"#s of #"NH"_3#
  2. From #"mol"#s of #"NH"_3# to #"mol"#s of #"NO"#
  3. From #"mol"#s of #"NO"# to #"g"# of #"NO"#

#"g NO"#

#= 9.6 cancel("g NH"_3) xx (cancel("1 mol NH"_3))/(17.0307 cancel"g NH"_3) xx (cancel"4 mol NO")/(cancel("4 mol NH"_3)) xx ("30.006 g NO")/(cancel"1 mol NO") = "Theoretical Yield"#

Notice how the units cancel out. This will give the theoretical yield based on the stoichiometry alone (the ideal result), and it is:

#color(green)("Theoretical Yield")# #=# #color(green)("16.9 g NO")#

Now, what you should do here is recall the equation for percent yield, which is really just asking, "how close did you get to perfect yield?".

#color(blue)("Percent Yield" = "Actual Yield"/"Theoretical Yield" xx 100%)#

#= ("12.0" cancel"g NO")/("16.9" cancel"g NO") xx 100%#

#=# #color(blue)("70.9% Yield of NO")#

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