JEE Mains · Chemistry · STD 11 - 9. Hydricarbon
In the following reaction The \(\%\) yield for reaction \(I\) is \(60 \%\) and that of reaction II is \(50 \%\). The overall yield of the complete reaction is \(..........\%\) [nearest integer]
- A \(31\)
- B \(32\)
- C \(33\)
- D \(30\)
Answer & Solution
Correct Answer
(D) \(30\)
Step-by-step Solution
Detailed explanation
Moles of benzene sulphonic acid before reaction \(II =0.6\,n\) Moles obtained for phenol (with \(\%\) yield \(=50 \%\) ) = \(0.6 \times 0.5\,n=0.3\,n\) So over all \(\%\) yield of complete reaction \(=\frac{0.3\,n }{ n } \times 100=30\)
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