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___________% 

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><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1753942485phphecXor.jpeg" alt="" width="415" height="139" loading="lazy"></picture></div><div>Let initial moles of reactant taken = n. Total moles obtained for benzene sulphonic acid = 0.6 n (with % yield = 60%)</div></div><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1753942531phpemURgM.jpeg" alt="" width="379" height="129" 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</p><p>reaction = <span class="mathml" contenteditable="false"> <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> </span>&lt;!- [if gte mso 9]>&lt;xml&gt; <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1815468167"> </o:OLEObject>&lt;/xml&gt;&lt;! [endif]-&gt;&nbsp;</p>

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