# Question e811d

Mar 3, 2017

${\text{0.43 moles N}}_{2}$

#### Explanation:

Start by writing a balanced chemical equation that describes this synthesis reaction

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

The thing to remember about balanced chemical equations is that the coefficients added in front of a chemical species represent the number of moles of said chemical species needed in order for the reaction to take place.

In this case, you need $1$ mole of hydrogen gas for every $\textcolor{b l u e}{3}$ moles of hydrogen gas in order to produce $\textcolor{red}{2}$ moles of ammonia.

Now, these mole ratios are true regardless of how many moles of a particular chemical species you have. In other words, you will have

${\text{number of moles of N"_2/"number of moles of NH}}_{3} = \frac{1}{\textcolor{red}{2}}$

Similarly, you will also have

${\text{number of moles of H"_2/"number of moles of N}}_{2} = \frac{\textcolor{b l u e}{3}}{1}$

and

${\text{number of moles of H"_2/"number of moles of NH}}_{3} = \frac{\textcolor{b l u e}{3}}{\textcolor{red}{2}}$

In your case, you know that $0.85$ moles of ammonia were produced by the reaction. Since the $1 : \textcolor{red}{2}$ mole ratio that exists between nitrogen gas and ammonia tells you that

${\text{no. of moles of N"_2 = 1/color(red)(2) * "no. of moles of NH}}_{3}$

you can say that the reaction consumed

"no. of moles of N"_2 = 1/color(red)(2) * "0.85 moles" = color(darkgreen)(ul(color(black)("0.43 moles N"_2)))#

The answer must be rounded to two sig figs, the number of sig figs you have for the number of moles of ammonia produced by the reaction.