Determining the quantity in moles of the gas that remains unreacted ?

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SUMMARY

The discussion centers on calculating the quantity of unreacted gas in a chemical reaction involving ammonia (NH3) and oxygen (O2) at 250 degrees Celsius and 800 atm. The reaction is represented as NH3(g) + 7/4 O2(g) --> NO2(g) + 3/2 H2O(g). Given 200 L of NH3 and 120 L of O2, the participant calculated that 131.4 L of NH3 remains unreacted, which translates to 2448.2 moles using the ideal gas law (n = PV/RT). The calculations and methodology presented are accurate and confirm the participant's understanding of the stoichiometry involved.

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Homework Statement



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.


Homework Equations





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.
 
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Seems OK.
 

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