# Under certain conditions of temperature and pressure, hydrogen gas and nitrogen gas react to form ammonia gas. Determine the mass of ammonia produced if 3.50 moles of hydrogen has reacts with 5.00 moles of nitrogen gas?

May 31, 2018

m("NH"_3)=39.7 color(white)(l) "g"

#### Explanation:

Start by balancing the equation for the reaction between hydrogen ${\text{H}}_{2}$ and nitrogen ${\text{N}}_{2}$, which produces ammonia ${\text{NH}}_{3}$:

$\textcolor{\mathrm{da} r k g r e e n}{1} \textcolor{w h i t e}{.} {\text{N"_2 (g) + color(darkgreen)(3) color(white)(.) "H"_2 to color(darkgreen)(2) color(white)(.) "NH}}_{3} \left(g\right)$

From the coefficients $\left(n \left({\text{N"_2))/(n("H}}_{2}\right)\right) = \frac{\textcolor{\mathrm{da} r k g r e e n}{1}}{\textcolor{\mathrm{da} r k g r e e n}{3}}$
meaning that all of the nitrogen and hydrogen molecules available would have been converted to ammonia if supplied at a $1 : 3$ ratio; otherwise one of the reagents will be in excess.

The question states that, however,

$\left(n ' \left({\text{N"_2))/( n' ("H}}_{2}\right)\right) = \frac{3.5}{5.0} = \frac{7}{10}$;

$\frac{7}{10} \textcolor{p u r p \le}{>} \frac{1}{3}$

meaning that the species represented in the numerator- nitrogen ${\text{N}}_{2}$- is in excess; The number of moles hence the mass of ammonia produced shall be calculated from the quantity of the limiting reactant- ${\text{H}}_{2}$- available. Also, from coefficients in the balanced equation:

$\left(n \left({\text{NH"_3))/(n("H}}_{2}\right)\right) = \frac{\textcolor{\mathrm{da} r k g r e e n}{2}}{\textcolor{\mathrm{da} r k g r e e n}{3}}$

$n \left({\text{NH"_3)=(color(darkgreen)(2))/(color(darkgreen)(3))*n("H}}_{2}\right)$
color(white)(n("NH"_3))=(color(darkgreen)(2))/(color(darkgreen)(3))*3.50 color(white)(l) "mol"
color(white)(n("NH"_3))=2.33 color(white)(l) "mol"

Take $17.031 \textcolor{w h i t e}{l} g \cdot {\text{mol}}^{- 1}$ as the molar mass $M$ of ammonia ${\text{NH}}_{3}$,

$m \left({\text{NH"_3)=n("NH"_3)*M("NH}}_{3}\right)$
color(white)(m("NH"_3))=2.33 color(white)(l) color(red)(cancel(color(black)("mol"))) * 17.031 color(white)(l) g*color(red)(cancel(color(black)("mol"^(-1))))
color(white)(m("NH"_3))=39.7 color(white)(l) "g"