# 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
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 $\text{NO}$ based on the amount of ${\text{NH}}_{3}$---the limiting reagent---that you have.

$\textcolor{g r e e n}{\text{Actual Yield}}$ $=$ $\textcolor{g r e e n}{\text{12.0 g NO}}$

Based on the reaction stoichiometry, let's see how much of $\text{NO}$ product you SHOULD have made.

The mass of product $\text{NO}$ can be calculated by doing the following conversions in sequence:

1. From $\text{g}$ of ${\text{NH}}_{3}$ to $\text{mol}$s of ${\text{NH}}_{3}$
2. From $\text{mol}$s of ${\text{NH}}_{3}$ to $\text{mol}$s of $\text{NO}$
3. From $\text{mol}$s of $\text{NO}$ to $\text{g}$ of $\text{NO}$

$\text{g NO}$

= 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:

$\textcolor{g r e e n}{\text{Theoretical Yield}}$ $=$ $\textcolor{g r e e n}{\text{16.9 g NO}}$

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%

$=$ $\textcolor{b l u e}{\text{70.9% Yield of NO}}$