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# One Mole of any gas occupies 22.4 L at STP

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Consider the following reaction, which takes place in an autoclave at 250C and 800 atm.

4NH3(g) + 7O2(g) = 4 NO2(g) + 6H2O(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 the gas that remains unreacted.

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#### Solution Preview

4 NH3 + 7O2 = 4 NO2 + 6 H2O

7 litre of O2 reqires 4 litres of NH3
120 litre of O2 rqiures (4/7)*120 L of NH3
120 L of O2 requires 480/7 L of NH3

200 - 480/7
(1400 - ...

#### Solution Summary

First of all make sure that the equation representing chemical reaction is balanced.
A balanced equation can guide us to calculate the volume of gas that remains unreacted.
Volume of left over gas must be found at Standard Temperature (273 K) and Pressure(1 Atmospheric Pressure)
Use the fact that one mole of any gas occupies 22.4 L volime at STP to calculate the number of moles in the left over gas.

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