1.

How much propanol is required for dehydration to get `2.24` litre of propene at `N.T.P`. If yield is `100%`

Answer» `C_(3)H_(8)O+H_(2)SO_(4)rarrC_(3)H_(6)+H_(2)O+H_(2)SO_(4)`
Molecular weight of propanol `=60`
from the equation given above we can see that from dehydration of `1` mole or `60` gram of propanol we get `1` mole (`22.4`lit) of propene as product.
`because 22.4` litre of `C_(3)H_(6)` can be get from dehydration of `60g` of propanol.
`therefore 1` litre of propene can be get from dehydration of `(60)/(22.4)g` of propanol


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