Question

Given the reaction, 3H2+N2----> 2NH3. How many L of NH3 are formed at 85C and 1.25atm from 30.6L of H2 at 20C and 750 torr and 10.8g of N2?

Answer 1

Number of moles of H2 will be = pV / RT

750 torr = 0.986842 atm

20C = 293.15 K

R = 0.0821 L.atm/mol.K

So putting all the values we get :

=(0.986842 x 30.6) / (0.0821 x 293.15)

= 1.25 moles

Number of moles of N2 = 10.8 / 28 = 0.386 moles

So clearly hydrogen is the limiting reagent

3 moles of H2 makes 2 moles of NH3

So 1.25 moles of H2 will make ( 1.25 x 2) / 3 = 0.83 moles of ammonia

The volume of 0.83 moles ammonia will be :

V = nRT / p

T = 85C = 358.15 K

V = (0.83 x 0.0821 x 358.15) / 1.25

V = 19.5 L

So 19.5 L of NH3 will be formed.