When 32.25 g of ethyl chloride is subjected to dehydrohalogenation reaction, the yield of alkene formed is 50%. The mass of the product formed is (at

Correct Answer - B
`underset("Ethyl chloride")(H-underset(H)underset(|)overset(H)overset(|)(C)-underset(H)underset(|)overset(Cl)overset(|)(C)-H)+KOHtoKCl+H_(2)O+H-underset("Ethene")(underset(H)underset(|)(C)=underset(H)underset(|)(C))-H`
Molar mass of ethyl chloride
`= 2xx12+5xx1+1xx35.5`
`=24+9+35.5=64.5gmol^(-1)`
Molar mass of ethene `=2+12+4xx1`
`=28gmol^(-1)`
Actual yield = 50% of 28 g = 14g
64.5g of ethyl chloride on dehydrohalogenation gives 14 g ethene
`:.` 32.25 g of ethyl chloride on dehydrohalogenation will give ethene = 7g