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r3zbabii217 · Jul 25, 2009, 12:23 AM

Consider the following reaction, which takes place in an autoclave at 250^oC and 800 atm.
Na3(g)+ 7/4O2(g) -> NO2(g)+ 3/2H2O(g)
Into the reaction vessel has been placed 200L of NH3(g) and 120L of O2(g). The reaction is allowed to go to completion. Determine the quantity, in moles, of gas that remains unreacted.

Unknown008 · Jul 26, 2009, 09:12 AM

I hope you mean:

$NH_3(g)+ \frac{7}{4}O_2(g) \rightarrow NO_2(g)+ \frac32 H_2O(g)$

Find the volume ratio of ammonia to oxygen first. You should have 1 volume of ammonia reacting with 7/4 volume of oxygen.

Now, you have 200 L of ammonia. That would react with 200*(7/4) L of oxygen, that is 350 L of oxygen.

But you only have 120 L. That means that you;ll have 120*(4/7) L of ammonia, that is 68.6 L.

So, you will have 131.4 L of excess ammonia.

Now, one mole of a gas at standard conditions occupies 22.4 L.
You can find the number of moles of the unreacted gas.