Question

When 20 g of Sulphur dioxide reacts with oxygen, 23 g of Sulphur trioxide is formed. What is the percentage yield?​

Answer 1

Answer:

80 percent is the percentage yielf

Answer 2

Answer:

The percentage of yield obtained is 92%

Explanation:

The reaction is as follows:

$SO_{2} + O_{2} -- > SO_{3}$

20g                    23g

Molar mass of Sulphur-di-oxide = 32 + (16x2) = 64g

Molar mass of Sulphur-tri-oxide = 32 + (16x3) = 80g

For 64g of Sulphur-di-oxide = 80g of Sulphur-tri-oxide is obtained.

So, for 20g of Sulphur-di-oxide = x g of Sulphur-tri-oxide is obtained.

Therefore, on finding the value of the actual yield of Sulphur-tri-oxide,

x = $\frac{80 * 20}{64}$

x = 25g

The theoretical yield is 25g

The actual yield is 23g. To calculate the yield percentage,

Yield percentage = $\frac{Actual yield}{Theoretical yield} * 100$

                             = $\frac{23}{25} * 100$

                             = 92%

Therefore, the percentage yield is 92%.

In this way, percentage yield can be calculated accurately for any chemical equation.

#SPJ2