In the following reaction ​​​​​​​ The % yield for reaction I is 60 % and that of reaction II is 50 % . The…

Let initial moles of reactant taken $=\mathrm{n}$
Total moles obtained for benzene sulphonic acid (with $\%$ yield $=60\%$) $=0.6 \mathrm{n}$

Moles of benzene sulphonic acid before reaction $\mathrm{II}=0.6 \mathrm{n}$

Moles obtained for phenol (with $\%$ yield = $50\%$) $=0.6\times 0.5 \mathrm{n}=0.3 \mathrm{n}$

So over all $\%$ yield of complete reaction $=\frac{0.3\mathrm{n}}{\mathrm{n}}\times 100=30$