# Stoichiometry: efficiency

My book gives me the following problem: "A mixture of $125.0 g$ $\ce{N2}$ and $32.0 g$ $\ce{H2}$ reacts to $36.5 g$ $\ce{NH3}$. Calculate the efficiency."

My method to do this is - one I learned from my teacher, to put every given molecule's amount of substance (moles) in a table with 3 steps: start (S), gone (G)/formed (F) and rest (R).

$125g\:N_2 \leftrightarrow n=4.462$ moles

$32g\:\left(3\right)H_2\leftrightarrow n=16.13$ moles

\begin{matrix} &N_2&+&3H_2&\longrightarrow &2NH_3\\S&4.462&&16.13&&\\G&4.462&&13.386&F&8.924\\R&0&&2.74&&8.924\end{matrix}

($\rightarrow$ abundance of $H_2$)

Now, what is here the efficiency? Is it $\frac{36.5\:g}{8.924\:moles\cdot M_{NH_3}}\approx 24\%$?

• I don't know what is wrong with your calculator, but when I divide 32.0 by 2.00, I get 16.0, not 16.13. – LDC3 Apr 5 '14 at 15:13
• @LDC3 true, while $M_{\ce{H2}}\approx2.02$ g/mol and therefore it should rather be $15.84$ mol. – Martin - マーチン Apr 5 '14 at 16:03

If you look at the chemical reaction: $$\ce{N2 + 3H2 \rightarrow 2 NH3}$$
You can see that you can, at best, make 2 mole of $\ce{NH3}$ per mole of $\ce{N2}$ (for the case here where $\ce{H2}$ is abundant). Given that you have $4.46\; moles$ of $\ce{N2}$ you could at best make $8.92\; moles$ of $\ce{NH3}$ which would be $151.9\; g$. Since you have made only $36.5\;g$ that means that your efficiency is $36.5/151.9=24\%$
• The formula weight of $NH_3$ is 17, not 16. – LDC3 Apr 5 '14 at 15:16