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Grade: 12th pass
        60ml of a mixture of Nitrous oxide and nitric oxide was exploded with excess of hydrogen. If 38 ml of N2was formed , calculate the volume of each gas in the mixture.
3 years ago

Answers : (2)

vaibhav
14 Points
							
let x be the volume of NO 
the reaction takes as 
2NO +2H2 ----> N2 +2H2O
2x                    x
N2O +H2 ----->N2 +H2O
60-2x              60-2x
the volume of notrogen collected is 38
therefore   60-2x+x=38
                x=44
hence volume of NO and N2O are 44 and 16 respectively
one month ago
Vikas TU
9205 Points
							
Dear student 
If we take the volume of the mixture to be 2x ml of NO. Then we get that the  volume of N2O will be present = (60−2x) ml.
We can write the balanced chemical equations as such:
Equation-               2NO + 2H2 -> N2+2H2O  .
Number of moles -   2x                    x.
Equation-                N2O + H2→ N2+H2O.
Number of moles-  60−2x         60−2x.
Thus from the quetion we get to know that the volume of N2 collected is 38 ml.
Hence, we can write that 60−2x+x=38 or x=22 ml.
Therefore, the volumes of NO and N2O which are present in the mixture will be:
The volume of NO will be=2x=2×22=44 ml.
The volume of N2O will be=60−44−16 m.
Regards  
one month ago
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