# What is the mass of ammonia formed if the reaction between 14 g of nitrogen and 12 g of ammonia gives a 70 % yield?

Aug 14, 2016

The mass of ammonia will be 13 g.

#### Explanation:

We must first identify the limiting reactant, and then we calculate the theoretical yield.

$\textcolor{w h i t e}{m m m m m m m l} {\text{ N"_2 +color(white)(l) "3H"_2 → color(white)(ll)"2NH}}_{3}$

$\text{MM/g·mol"^"-1": " } 28.01 \textcolor{w h i t e}{l l} 2.016 \textcolor{w h i t e}{m m l} 17.03$

(a) Identify the limiting reactant

We calculate the amount of ${\text{NH}}_{3}$ that can form from each reactant.

Calculate the moles of ${\text{N}}_{2}$.

${\text{Moles of N"_2 = 14 color(red)(cancel(color(black)("g N"_2))) × "1 mol N"_2/(28.01 color(red)(cancel(color(black)("g N"_2)))) = "0.500 mol N}}_{2}$

Calculate moles of ${\text{NH}}_{3}$ formed from the ${\text{N}}_{2}$

0.500color(red)(cancel(color(black)("mol N"_2))) × "2 mol NH"_3/(1 color(red)(cancel(color(black)("mol N"_2)))) = "1.00 mol NH"_3

Calculate the moles of ${\text{H}}_{2}$

${\text{Moles of H"_2 = 12 color(red)(cancel(color(black)("g H"_2))) × "1 mol H"_2/(2.016 color(red)(cancel(color(black)("g H"_2)))) = "5.95 mol H}}_{2}$

Calculate moles of ${\text{NH}}_{3}$ formed from the ${\text{H}}_{2}$

5.95 color(red)(cancel(color(black)("mol H"_2))) × "2 mol NH"_3/(3 color(red)(cancel(color(black)("mol H"_2)))) = "3.97 mol NH"_3

The limiting reactant is ${\text{N}}_{2}$, because it produces fewer moles of ${\text{NH}}_{3}$.

(b) Calculate the theoretical yield of ${\text{NH}}_{3}$.

${\text{Theoretical yield" = 1.00 color(red)(cancel(color(black)("mol NH"_3))) × "17.01 g NH"_3/(1 color(red)(cancel(color(black)("mol NH"_3)))) = "17.0 g NH}}_{3}$

(c) Calculate the actual yield of ${\text{NH}}_{3}$

If % yield = 75 %, then

$\text{Actual yield" = 17.0 color(red)(cancel(color(black)("g theoretical"))) × "75 g actual"/(100 color(red)(cancel(color(black)("g theoretical")))) = "13 g actual}$