# Equilibrium in gas reaction?

At a pressure of 2.00 × 107 Pa, 1.00 moles of nitrogen was mixed with 3.00 moles of hydrogen. The final equilibrium mixture formed contained 0.300mol of ammonia.

Calculate the amounts, in mol, of nitrogen gas and hydrogen gas in the equilibrium mixture.

The answers are 0.850 mol for nitrogen gas and 2.55mol for hydrogen gas but I can't get the asnwer.

This is what I had done:

N2 + 3 H2 <--> 2 NH3

........Initial.....Change......Equilibrium

N2.....1...........+(-0.30)..........0.7 mol

H2.....3..........+3(-0.3).............2.1 mol

NH3...0...........-(-0.3)......... 0.3 mol

Are there any mistakes?

### 1 Answer

- hcbiochemLv 73 months agoFavorite Answer
Yes, there are mistakes. You haven't correctly taken into account the stoichiometry of the reaction.

Moles H2 consumed = 0.30 mol NH3 X (3 mol H2 / 2 mol NH3) = 0.45 mol H2 consumed

Moles H2 remaining = 3.00 - 0.45 = 2.55 mol H2

Moles N2 consumed = 0.30 mol NH3 X (1 mol N2 / 2 mol NH3) = 0.15 mol N2 consumed

moles N2 remaining = 1.00 - 0.15 = 0.85 mol N2

If you want to do the ICE table correctly:

2x = 0.30 mol, so x = 0.15 mol

........Initial.....Change......Equilibrium

N2.....1...........-x..........1-x = 0.85 mol

H2.....3..........-3x............3 - 3x = 2.55 mol

NH3...0...........+2x......... 0+2x = 0.3 mol