A) 5.15g
B) 2.56g
C) 7.49g
D) 5.80g
Correct Answer: A
Solution :
Initial mole of butane \[=\frac{5.8}{58}=0.1\] \[\underset{0.1-x}{\mathop{Bu\tan e}}\,\rightleftharpoons \underset{x}{\mathop{Isobu\tan e}}\,\] \[{{K}_{C}}=8=\frac{x}{0.1-x}\] or \[x=0.088\] m(isobutene) \[=0.088\times 58=5.15g\]You need to login to perform this action.
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