Question

Ammonia may be oxidised to nitrogen monoxide in the presence of a catalyst according to the following equation. 4NH3 +5O2-4NO+6H2O.If 27 litres of reactants are consumed ,what volume of nitrogen monoxide is produced at the same temperature and pressure?
NO is 2/5 of the output of 10 moles.

Since the original 9 moles occupied 27 liters, the 10 moles occupy 30 liters.

So, NO occupies 2/5 * 30 = 12 liters.

This assumes that the H2O is also a vapor.
Total volume of reactants will be 9 moles.
Volume of NO is 4.
When 9 moles of reactants give4 volume of NO then 27liter of reactants will produce what volume of NO.

27*4/9
12LITER of NO.

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