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

($\rightarrow$ abundance of $H_2$)

Now, what is here the efficiency? Is it $\frac{36.5\:g}{8.924\:moles\cdot M_{NH_3}}\approx 24\%$?
Yes, that is completely correct. Let me summarize the approach.

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 $36.5/151.9=24\%$