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]

Let initial moles of reactant taken = n. Total moles obtained for benzene sulphonic acid = 0.6 n (with % yield = 60%)

Moles of benzene sulphonic acid before reaction II = 0.6n

Moles obtained for phenol (with % yield 50%) = 0.6 × 0.5 n = 0.3 n

So, overall % age yield of complete reaction = $\frac{0.3n}{n}\times 100=30\mathrm{\%}$

<div><div><picture><img src="https://images.shiksha.com/mediadata/images/1753607769phppTdi1l.jpeg" alt="" width="298" height="100" loading="lazy"></picture></div></div><p>Let initial moles of reactant taken = n. Total moles obtained for benzene sulphonic acid = 0.6 n (with % yield = 60%)</p><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1753607786phpMxej73.jpeg" alt="" width="268" height="123" loading="lazy"></picture></div></div><p>Moles of benzene sulphonic acid before reaction II = 0.6n</p><p>Moles obtained for phenol (with % yield 50%) = 0.6 × 0.5 n = 0.3 n</p><p>So, overall % age yield of complete&nbsp;reaction =&nbsp;<math><mrow><mfrac><mrow><mn>0</mn><mo>.</mo><mn>3</mn><mi>n</mi></mrow><mrow><mi>n</mi></mrow></mfrac><mo>×</mo><mn>1</mn><mn>0</mn><mn>0</mn><mo>=</mo><mn>3</mn><mn>0</mn><mn>%</mn></mrow></math></p>

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