# Using the equation "N"_2 + 3"H"_2 -> 2"NH"_3, if 28g "N"_2 react, how many grams of "NH"_3 will be produced?

Jul 1, 2016

${\text{34 g NH}}_{3}$

#### Explanation:

The key to any stoichiometry problem is the mole ratio that exists in the balanced chemical equation between the two chemical species that are of interest.

In this case, you have

${\text{N"_ (2(g)) + 3"H"_ (2(g)) -> color(red)(2)"NH}}_{3 \left(g\right)}$

The balanced chemical equation tells you that the reaction consumes $3$ moles of hydrogen gas, ${\text{H}}_{2}$, and produces $\textcolor{red}{2}$ moles of ammonia, ${\text{NH}}_{3}$, for every mole of nitrogen gas, ${\text{N}}_{2}$, that takes part in the reaction.

Since the problem doesn't mention the amount of hydrogen gas available for the reaction, you can assume that it's in excess, which means that the reaction will consume all the grams of nitrogen gas present.

Now, you can convert the mole ratio to a gram ratio by using the molar masses of the two species

M_("M N"_2) ~~ "28 g mol"^(-1)

M_("M NH"_3) ~~ "17 g mol"^(-1)

This means that the 1:color(red0(2) mole ratio that exists between nitrogen gas and ammonia can be written as

$\left(1 \textcolor{g r a y}{\cancel{\textcolor{b l a c k}{\text{mole"))) * "28 g" color(gray)(cancel(color(black)("mol"^(-1)))))/(color(red)(2) color(gray)(cancel(color(black)("moles"))) * "17 g" color(gray)(cancel(color(black)("mol"^(-1))))) = (28 color(gray)(cancel(color(black)("g"))))/(34color(gray)(cancel(color(black)("g}}}}\right) = \frac{14}{17} \to$ gram ratio

So, if the reaction consumes $\text{28 g}$ of nitrogen gas, it follows that it will produce

28 color(red)(cancel(color(black)("g N"_2))) * "17 g NH"_3/(14color(red)(cancel(color(black)("g N"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("34 g NH"_3)color(white)(a/a)|)))

In other words, $1$ mole of nitrogen gas will produce $\textcolor{red}{2}$ moles of ammonia.

The answer is rounded to two sig figs.