[Evaldas]

Nitrogen and hydrogen gases are mixed in ratio 1:3. 100 liters of such gas mixture had been let through catalyst and after the reaction the volume of the mixture decreased by 40%. Calculate the composition of the gas mixture after the reaction in percents of the volume.

Ok: the reaction is N₂+3H₂ ::equil::2NH₃
Also 25 l of hydrogen and 75 l of nitrogen reacted. After the reaction there were 60 l of the gas mixture, which now contains nitrogen, hydrogen and ammonia. And if I'm getting it right in those 60l there are x l of hydrogen, 3x l of nitrogen, and y l of ammonia. So the first equation would be 4x+y=60. What about the other? I don't understand.

[Borek]

If y liters of ammonia were produced, how many liters of nitrogen were consumed?

[Evaldas]

1.5y?
So can I write it as y/3+1.5y+y=60?

[Borek]

1.5y?

No?

[Evaldas]

Sorry, that's for hydrogen...
0.5y then?

[Evaldas]

Ok, I still don't understand and I still need help 

[Borek]

Have you tried 0.5y?

Don't wait to be pushed on every step, try to approach the question by yourself.

[Evaldas]

Have you tried 0.5y?

Don't wait to be pushed on every step, try to approach the question by yourself.

Ok... My newest idea: 75-1.5y+25-0.5y+y=60, y=40
Meaning, 40l of ammonia were produced, 5l of Nitrogen left, 15l of hydrogen left (40+15+5=60l).
So then the mixture contains ~66.7% ammonia, ~8.3% nitrogen, 25% hydrogen. 
I'm out of ideas

[Rutherford]

Correct.