# Question 89a6e

Jan 31, 2018

$a . {N}_{2} \left(g\right) + 3 {H}_{2} \left(g\right) \to 2 N {H}_{3} \left(g\right)$
$b . \approx 577 g N {H}_{3}$
c. ~~69.0% yield

#### Explanation:

1. Write and balance the given chemical equation.
color(red)(N_2+3H_2.->2NH_3

2. Find the required molar masses of the involved substances.
${N}_{2} = 28.0 \text{ g/mol}$
$N {H}_{3} = 17.0 \text{ g/mol}$

3. Given the masses of both raw material and the production output, theoretical yield of this production run can be computed as follows:
$= 475 \cancel{g {N}_{2}} \times \frac{1 m o l {N}_{2}}{28.0 \cancel{g {N}_{2}}}$
$= 16.96 m o l {N}_{2} \approx 17.0 m o l {N}_{2}$

Now, convert mole ${O}_{2}$ $\left(\eta {O}_{2}\right)$ to $\eta N {H}_{3}$ using the molar ratio of these two(2) substances that is obtainable from the balanced equation described in $\underline{\text{step } 1}$.
$= 17.0 \cancel{m o l {N}_{2}} \times \frac{2 m o l N {H}_{3}}{1 \cancel{m o l {N}_{2}}}$
$= 33.93 m o l N {H}_{3} \approx 34.0 m o l N {H}_{3}$

Then, find the $\text{theoretical yield}$. Knowing the molar mass of $N {H}_{3}$, conversion factor is obtainable and the production output can be computed. Make sure units work out and the desired unit is attained; i.e.,
color(red)(=34.0cancel((molNH_3)xx(17.0gNH_3)/(1cancel(molNH_3))
$\textcolor{red}{= 576.79 g N {H}_{3} \approx 577 g N {H}_{3}}$

1. Finally, find the percent yield in this production run; i.e.,
% yield=("actual yield")/("theoretical yield")xx100
% yield=(397cancel(gNH_3))/(577cancel(gNH_3))xx100
color(red)(% yield=68.8%~~69.0% #