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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.
let x be the volume of NO the reaction takes as 2NO +2H2 ----> N2 +2H2O2x xN2O +H2 ----->N2 +H2O60-2x 60-2xthe volume of notrogen collected is 38therefore 60-2x+x=38 x=44hence volume of NO and N2O are 44 and 16 respectively
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
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