Question

Ammonia can also be synthesized by the reaction:
3 H2(g) + N2(g) S2 NH3(g)
What is the theoretical yield of ammonia, in kg, that we can synthesize from 5.22 kg
of H2 and 31.5 kg of N2 ?

Answer 1

The main equation we use here is
no. of moles (n) = mass (m) divided by molar mass (M)
since we are given with the masses of H2 and N2 we will 1st find the no. of moles of each.
HYDROGEN--------->Mass= 5.22kg molar mass = 1gmol^-1= 0.001kgmol
so no. of moles = 5.22kg /0.001kgmol^-1 = 5220moles

NITROGEN---------> Mass= 31.5kg Molar mass = 14gmol^-1 = 0.014kgmol^-1
So no. of moles (n) = 31.5kg /0.014kgmol^-1 = 2250 moles

Now look at the reacting ratio of hydrogen to nitrogen tht is 3:1 right?
To react with 2250 moles of nitrogen we will need 2250*3 moles of hydrogen right ?
So the no. of moles of hydrogen to react with 2250 moles of nitrogen is 6750 moles. But we have only 5220 moles of hydrogen right? So the limiting factor is Hydrogen.

Since we have a limiting agent here we will compare the mole ratio of NH3 with that.
When 3 moles of hydrogen is reacted 2 moles of ammonia is formed . We just have to find the moles formed if 5220 moles of hydrogen react right?
if 3 moles is reacted to form 2 moles of NH3
one mole is reacted to form 2/3 moles
if 5220 moles are reacted the no of moles of nh3 formed is
\[5220 * \frac{ 2 }{ 3}= 3480 moles\]
now we know the moles of NH3 , molar mass of NH3 (0.017kgmol^-1)
use m = n*M
m = 3480mol * 0.017kgmol^-1 = 59.16kg of nh3