# If the reaction of 77.0 grams of CaCN2 produces 27.1 grams of NH3, what is the percent yield? CaCN2+3H2O=>CaCO3+2NH3

Dec 12, 2014

The answer is 83.3%.

From the balanced chemical equation

$C a C {N}_{2} + 3 {H}_{2} O \to C a C {O}_{3} + 2 N {H}_{3}$

we can see that we have a $1 : 2$ mole ratio between $C a C {N}_{2}$ and $N {H}_{3}$; that is, for every mole of $C a C {N}_{2}$ used in the reaction, 2 moles of $N {H}_{3}$ will be formed.

In order to determine the percent yield, we must determine what the limiting reagent is. We know that $C a C {N}_{2}$ and $N {H}_{3}$'s molar masses are $80 \frac{g}{m o l}$ and $17 \frac{g}{m o l}$, respectively, therefore

${n}_{C a C {N}_{2}} = {m}_{C a C {N}_{2}} / \left(m o l a r m a s s\right) = \frac{77 g}{80.0 \frac{g}{m o l}} = 0.96$ moles and

${n}_{N {H}_{3}} = {m}_{N {H}_{3}} / \left(m o l a r m a s s\right) = \frac{27.1 g}{17 \frac{g}{m o l}} = 1.6$ moles

According to our mole-to-mole ratio, $0.96$ moles of $C a C {N}_{2}$ should have produced $2 \cdot 0.96 = 1.92$ moles of $N {H}_{3}$, but instead produced $1.6$ moles. This means that not all $C a C {N}_{2}$ reacted <=> water is the limiting reagent.

So, the percent yield, defined as the actual yield divided by the theoretical yield and multiplied by 100%, is

%yield = (actual)/(the o re tic) * 100% = (1.6 mol es)/(1.92 mol es) * 100% = 83.3%