# 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"?

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%

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