# How many grams of nitrogen are necessary to produce 15.00 grams of ammonia?

Jan 14, 2017

The required mass of nitrogen is 12.34 g.

#### Explanation:

The balanced equation is:

$0.5 {\text{N"_2 + 1.5"H"_2 rarr "NH}}_{3} \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \left(1\right)$

The molecular mass of ammonia, ${\text{NH}}_{3}$, is equal to 14.01 g + 3(1.008 g), or 17.03 g.

$\text{Moles of ammonia" = "mass in grams"/"molecular mass" ="15.00 g"/"17.03 g/mol" = 15.00/17.03 " mol}$

Therefore, according to equation $\left(1\right)$,

$\text{Moles of nitrogen" = 0.5(15.00/17.03) " mol} \textcolor{w h i t e}{\ldots \ldots \ldots \ldots} \left(2\right)$

The number of moles of nitrogen is calculated by the following equation:

"Moles of N"_2 = ("mass of nitrogen")/ ("molar mass of nitrogen")

Thus,

${\text{Mass of nitrogen" = "moles of N"_2*"molar mass of N}}_{2} \textcolor{w h i t e}{\ldots} \left(3\right)$

Therefore, from equations $\left(2\right)$ and $\left(3\right)$, we have:

$\text{Mass of nitrogen" = 0.5(15.00/17.03) " mol" xx "28.02 g/mol" = "12.34 g}$

With kind regards
Dr. Mamdouh Younes