Question

In a certain experiment, 1.000 mole of N2 (g) and 1.000 mole of H2 were allowed to react at 500°C in a 1.000-liter flask. The relevant reaction was as follows: N2 (g) + 3H2 (g) -> 2NH3 (g)
After the system reached equilibrium, the flask was found to contain 0.921 mole of N2 (g). Calculate the equilibrium concentrations of H2(g) and NH3(g).

Answer 1

Initial (N2) = 1 mo/L
Initial (H2) = 1 mol/L
........................N2 (g) + 3H2 (g) -> 2NH3 (g)
I.....................1.0............1.0................0
Change...........-x.............-3x..............2x
equil...............0.921..............................
change for N2 must be 1.000 - 0.921= 0.079 = x
Then 3x must be 3*0.079 = 0.237 and 1 - 3x will be ?
and 2x must be 2*0.079 = 0.158 and 0 + 2x will be ?
Post your work if you get stuck.

Answer 2

Okay, thank you! That makes sense now.