What mass of isobutylene is obtained from 37g of tertiary butyl alcohol by heating with 20% `H_(2)SO_(4)` at 363 K if the yield is 65% ?

Options

A. 16 g
B. 18 g
C. 20 g
D. 22 g

Answer

B

Explanation

`4 xx 12 + 16 + 10 = 74 g " " 4 xx 12 +8 = 56g`
74 g of t-BuOH gives Isobutylene = 56 g
37g of t-BuOH gives Isobutylene
`= (56 xx 37)/(74) =28 g`
But the yield of product = 65%
`:.` Actual Isobutylene formed
`= (28 xx 65)/(100) = 18.20 g`