# How many grams of NH_3 can be produced from the reaction of 28 g of N_2 and 25 g of H_2?

##### 1 Answer
Jan 12, 2016

Well, what is the reaction itself? You can always start there.

${N}_{2} \left(g\right) + {H}_{2} \left(g\right) r i g h t \le f t h a r p \infty n s N {H}_{3} \left(g\right)$

Now balance it. You have two reactants with compositions involving a single element, which makes it very easy to keep track of how much is on each side. I would balance the nitrogens, and then the hydrogens.

(If you balance the hydrogen reactant with a whole number first, I can guarantee you that you will have to give ${\text{NH}}_{3}$ a new stoichiometric coefficient.)

${N}_{2} \left(g\right) + 3 {H}_{2} \left(g\right) r i g h t \le f t h a r p \infty n s 2 N {H}_{3} \left(g\right)$

The stoichiometric coefficients tell you that if we can somehow treat every component in the reaction as the same (like on a per-mol basis, hinthint), then one "[molar] equivalent" of nitrogen yields two [molar] equivalents of ammonia.

Luckily, one $\text{mol}$ of anything is equal in quantity to one $\text{mol}$ of anything else because the comparison is made in the units of $\text{mol}$s. So what do we do? Convert to $\text{mol}$s (remember the hint?).

${\text{28" cancel("g N"_2) xx ("1 mol N"_2)/((2xx14.007)cancel("g N"_2)) = "0.9995 mol N}}_{2}$
(how convenient... pretty much $\text{1 mol}$)

At this point you don't even need to calculate the number of $\setminus m a t h b f \left(\text{mol}\right)$s of $\setminus m a t h b f \left({\text{H}}_{2}\right)$. Why? Because ${\text{H}}_{2}$ is about $\text{2 g/mol}$, which means we have over $\text{10 mol}$s of ${\text{H}}_{2}$. We have ${\text{1 mol N}}_{2}$, and we need three times as many $\text{mol}$s of ${\text{H}}_{2}$ as we have ${\text{N}}_{2}$.

After doing the actual calculation you should realize that we have about $4$ times as much ${\text{H}}_{2}$ as we need. Therefore the limiting reagent is clearly ${\text{N}}_{2}$.

Thus, we should yield $2 \times 0.9995 = \text{1.9990 mol}$s of ${\text{NH}}_{3}$ (refer back to the reaction). So this is the second and last calculation we need to do:

"1.9990" cancel("mol NH"_3) xx ("17.0307 g NH"_3)/(cancel("1 mol NH"_3)) = color(blue)("34.0444 g NH"_3)