# If 156 g of sodium nitrate react, and 112 g of sodium nitrite are recovered, what is the percentage yield?

Jun 21, 2017

%"yield" = 88.4%

#### Explanation:

I'm going to assume the reaction is the following:

$2 {\text{NaNO"_3(s) rarr 2"NaNO"_2 (s) + "O}}_{2} \left(g\right)$

with an supplement of heat energy to "drive" the reaction.

We're given that $156$ ${\text{g NaNO}}_{3}$ react, so let's convert this to moles using its molar mass, calculated to be $85.00 \text{g"/"mol}$:

156cancel("g NaNO"_3)((1color(white)(l)"mol NaNO"_3)/(85.00cancel("g NaNO"_3))) = 1.84 ${\text{mol NaNO}}_{3}$

Now, using the stoichiometrically equivalent values (the coefficients), let's determine the relative number of moles of ${\text{NaNO}}_{2}$ that should form:

1.84cancel("mol NaNO"_3)((2color(white)(l)"mol NaNO"_2)/(2cancel("mol NaNO"_3))) = 1.84 ${\text{mol NaNO}}_{2}$

Finally, let's use the molar mass of sodium nitrite ($69.00 \text{g"/"mol}$) to calculate the theoretical mass yield:

1.84cancel("mol NaNO"_2)((69.00color(white)(l)"g NaNO"_2)/(1cancel("mol NaNO"_2))) = color(red)(127 color(red)("g NaNO"_2

We're given that the actual yield of ${\text{NaNO}}_{2}$ is $112$ $\text{g}$, so we can use the percent yield equation

%"yield" = "experimental yield"/"theoretical yield" xx 100%

to find the percent yield:

%"yield" = (112cancel("g NaNO"_2))/(color(red)(127)cancel(color(red)("g NaNO"_2))) xx 100% = color(blue)(88.4%

The percent yield of sodium nitrite for this reaction is thus color(blue)(88.4%.