1.

One litre of oxygen at NTP is allowed to reasonanace with three times of carbon monoxide at NTP. Calculate the volume of each gas found after the reaction.

Answer» The reaction equation is
`underset(2"vol")(2CO)+underset(1"vol")(O_(2)) to underset(2"vol")(2CO_(2))`
`"1 vol of" O_(2) "reacts with 2 vol. of CO"`
or `"1 litre of" O_(2) "reacts with 2 vol. of CO"`
`"Thus 1 litre of CO remains unchanged"`
`"1 vol. of" O_(2) "produces" CO_(2)=2"vol"`
`or "1 litre of" O_(2) "will produce" CO_(2)="2litre"`
Thus, gaseous mixture after the reaction consists.
Volume of CO=1litre
Volume of `CO_(2)`=2 litre


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