# A yield of NH_3 of approximately 98% can be obtained at 200 deg C and 1,000 atmospheres of pressure. How many grams of N_2 must react to form 1.7 grams of ammonia, NH_3?

May 26, 2017

1.429 g of ${N}_{2}$

#### Explanation:

We always have to read the problems completely to understand the question. In this case even when we have pressure and temperature conditions specified, they don't matter, first because they are constant and in second because the question ask for grams of ${N}_{2}$ to produce grams of $N {H}_{3}$.

We cannot compare grams of a substance to another in chemical reactions, we need to use moles, then we have to convert the weight of NH3 to moles.

$\eta = m / M W$
$M {W}_{N {H}_{3}} = 14.007 + 3 \left(1.008\right) = 17.031 g / m o l$
:. $\eta = 1.7 g / 17.031 g / m o l = 0.100 m o l$

The yield is 98%, then the number of moles that would be produced are:

$0.100 m o l / 0.98 = 0.102 m o l$ of $N {H}_{3}$

From the reaction (the hydrogen was implicit in the problem, because you are obtaining $N {H}_{3}$):

$N 2 + 3 H 2 \rightarrow 2 N H 3$

There are 2 parts of $N {H}_{3}$ that are obtained from 1 part of ${N}_{2}$

Then the moles of ${N}_{2}$ required are going to be the half of the $N {H}_{3}$ moles produced.

${N}_{2}$ moles = $N {H}_{3} \text{moles} / 2 = 0.102 m o l / 2 = 0.051 m o l$

Last step is too convert the moles back to grams using the MW of ${N}_{2}$

${m}_{{N}_{2}} = m o l \times M W = 0.051 m o l \times \left(14.007 g / m o l \times 2\right) = 1.429 g$