ans: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 %