Question

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

Answer 1

The mass of ammonia will be 13 g.

Explanation:

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

We start with the balanced equation.

`color(white)(mmmmmmml)" N"_2 +color(white)(l) "3H"_2 → color(white)(ll)"2NH"_3`

`"MM/g·mol"^"-1": " "28.01color(white)(ll) 2.016color(white)(mml) 17.03`

(a) Identify the limiting reactant

We calculate the amount of `"NH"_3` that can form from each reactant.

Calculate the moles of `"N"_2`.

`"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 `"NH"_3` formed from the `"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 `"H"_2`

`"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 `"NH"_3` formed from the `"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 `"N"_2`, because it produces fewer moles of `"NH"_3`.

(b) Calculate the theoretical yield of `"NH"_3`.

`"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 `"NH"_3`

If % yield = 75 %, then

`"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"`