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

[mcandrewsr]

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.

 

[Borek]

Seems OK.

 

[Blueshift5]

Yes, your approach is correct. To determine the quantity of unreacted gas in moles, you first need to determine the volume of each gas that has reacted. This can be done using the given volume ratio of 7/4 for O2 to NH3. This means that for every 7 moles of O2 that react, 4 moles of NH3 will also react.

In this case, we know that 120 L of O2 has reacted. Using the volume ratio, we can calculate that 68.6 L of NH3 has reacted (120 L x 4/7). This means that the remaining 200 L of NH3 (total amount placed in the reaction vessel) - 68.6 L (amount reacted) = 131.4 L of NH3 remains unreacted.

To convert this to moles, we use the ideal gas law, n = PV/RT, where P is pressure in atm, V is volume in L, R is the ideal gas constant (0.0821 L atm/mol K), and T is temperature in Kelvin. We are given the temperature and pressure, so we can plug in those values and solve for n.

n = (800 atm x 131.4 L) / (0.0821 L atm/mol K x 523.15 K) = 2448.1 mol NH3 unreacted.

Therefore, the quantity of unreacted gas in moles is 2448.1 mol NH3.