# Question #7d102

Mar 14, 2015

Your theoretical yield will be $\text{32.9 g}$ and the reaction's percent yield will be $\text{81.2%}$.

${N}_{2 \left(g\right)} + 3 {H}_{2 \left(g\right)} \to 2 N {H}_{3 \left(g\right)}$

Notice the $\text{3:2}$ mole ratio that exists between hydrogen gas and ammonia; this means that for every 3 moles of hydrogen gas that react, 2 moles of ammonia will be produced.

In other words, regardless of how many hydrogen moles react, you'll always have 2/3 less moles of ammonia produced.

Determine the number of moles of hydrogen gas by using its molar mass

${\text{5.84 g H"_2 * "1 mole H"_2/"2.016 g" = "2.897 moles H}}_{2}$

For a 100% yield, all the moles of hydrogen gas must react to produce ammonia. This means that the moles of ammonia produced will be

${\text{2.897 moles H"_2 * "2 moles NH"_3/"3 moles H"_2 = "1.931 moles NH}}_{3}$

Use ammonia's molar mass to see how many grams would be produced in a 100%-yield reaction $\to$ this is your theoretical yield.

${\text{1.931 moles NH"_3 * "17.03 g"/"1 mole NH"_3 = "32.9 g NH}}_{3}$

However, your reaction produces less ammonia (26.7 g) than what was calculated for a 100% yield, which means that your reaction's percent yield will be smaller than 100%.

$\text{% yield" = "actual yield"/"theoretical yield} \cdot 100$

$\text{% yield" = "26.7 g"/"32.9 g" * 100 = "81.2%}$