When a mixture consisting of 10 moles of SO 2 and 15 moles of O 2 was passed over a catalyst, 8 moles of SO 3 were formed. Thus, percentage yield of SO 3 wrt SO 2 is

Question

When a mixture consisting of 10 moles of SO 2 and 15 moles of O 2 was passed over a catalyst, 8 moles of SO 3 were formed. Thus, percentage yield of SO 3 wrt SO 2 is (A) 80 % (B) 100 % (C) 50 % (D) 25 %

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/83521486c954f55f549d0990f7b170d3-.png" /> If 10 mol SO 2 is taken, O 2 required is 5 mol, forming 10 mol SO 3 But SO 3 actual formed =8 mol Thus, percentage yield of SO 3 wrt SO 2=(8/10) × 100=80 %

Q. When a mixture consisting of $10$ moles of $S O_{2}$ and $15$ moles of $O_{2}$ was passed over a catalyst, $8$ moles of $S O_{3}$ were formed. Thus, percentage yield of $S O_{3}$ wrt $S O_{2}$ is

$80%$

$100%$

$50%$

$25%$

If $10 m o l S O_{2}$ is taken, $O_{2}$ required is $5 m o l$ , forming $10 m o l S O_{3}$
But $S O_{3}$ actual formed $= 8 m o l$
Thus, percentage yield of $S O_{3}$ wrt $S O_{2} = \frac{8}{10} \times$ $100 = 80%$

Q. When a mixture consisting of 10 moles of SO2​ and 15 moles of O2​ was passed over a catalyst, 8 moles of SO3​ were formed. Thus, percentage yield of SO3​ wrt SO2​ is

80%

100%

50%

25%