Question

When 14.5g of SO2 reacts with 21g of O2, what will be the theoretical yield and percentage yield of reaction if the actual yield is 12g?

Answer 1

We have to find theoretical yield and percentage yield of the reaction.

Solution : 14.5 g of SO2 reaches with 21 g of O2. and actual yield is 12g.

first we see chemical reaction is SO2 and O2,

2SO2 + O2 ⇒2SO3

we see, 2 moles of SO2 react with 1 mole of O2.

molecular weight of SO2 = 32 + 2 × 16 = 64 g

molecular weight of O2 = 32g

∴ 2 × 64 = 128g of SO2 reacts with 32g of O2.

∴ 14.5g of SO2 reacts with 32/128 × 14.5 = 3.625 g but given O2 is 21g

so, SO2 is limiting reagent.

hence, reaction prefers SO2 to produce SO3

2 moles of SO2 give 2 moles of SO3

∴ 128g of SO2 gives 2 × (32 + 3 × 16) = 160g of SO3

⇒14.5 g of SO2 gives 160/128 × 14.5 g = 18.125 g

so theoretical yield = 18.125 g

but actual yield = 12g

so, percentage yield = actual yield/theoretical yield × 100

= 12/18.125 × 100

= 66.207 %

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