help!

[shen]

Consider the following reactions which takes place in an autoclave at 250 degrees and 800 atm.

NH₃ (g) + 7/4 O₂(G) = NO₂ (g) + 3/2 H₂O (g)
Into the reaction vessel has been placed 200 L of NH₃ (g) and 120 L of O₂ (g). The reaction is allowed to go to completion. Determine the quantity, in moles, of the gas that remains unreacted.

[peter]

Consider the following reactions which takes place in an autoclave at 250 degrees and 800 atm.

NH₃ (g) + 7/4 O₂(G) = NO₂ (g) + 3/2 H₂O (g)
Into the reaction vessel has been placed 200 L of NH₃ (g) and 120 L of O₂ (g). The reaction is allowed to go to completion. Determine the quantity, in moles, of the gas that remains unreacted.

Fist you need to determine which is the limiting reagent?
200/1 > 120 / (7/4) => O₂ is limiting
Then calculate the amount of the NH₃ that has been reacted:
120 / (7/4) * 1 = 68.6 L
The volume that has been left 200-68.6 = 131.4
Assuming that you've meant 250 C, y the number of moles 131.4*800/(250+273)/0.082 = 2451 mol