1. The problem statement, all variables and given/known data Consider the following reaction, which takes place in an autoclave at 250 degrees C and 800 atm. NH3(g) + 7/4 O2(g) --> NO2(g) + 3/2 H20(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. 2. Relevant equations 3. The attempt at a solution I really don't know what to do, so I just tried this: The ratio of moles of O2 to NH3 is 7/4:1. Therefore the volume ratio is also 7/4:1. If you react 120 L of O2, then: 120 x 4/7 = 68.6 L NH3 200 L NH3 - 68.6 L NH3 = 131.4 L NH3 left unreacted Then I put it into n = PV/RT to convert to moles and I get 2448.2 mol NH3 left unreacted. IS THIS CORRECT??? Thanks Detailed explanations would be greatly appreciated.