Question
When 14.5g of so2 react with 21g of O2 what will be the theoretical and percentage yiad of the reaction if the actealyid of the reaction is 12g
Answers
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The balanced chemical equation for the reaction between SO2 and O2 is:
2 SO2 + O2 -> 2 SO3
From the equation, we can see that 2 moles of SO2 react with 1 mole of O2 to produce 2 moles of SO3.
First, we need to calculate the limiting reagent to determine the theoretical yield.
The moles of SO2 present = 14.5 g / 64.06 g/mol = 0.226 moles
The moles of O2 present = 21 g / 32 g/mol = 0.656 moles
From the above values, we see that O2 is in excess as it has more moles than required for the reaction. Therefore, SO2 is the limiting reagent.
Theoretical yield = (0.226 mol SO2) x (2 mol SO3/2 mol SO2) x (80.06 g/mol SO3) = 18.1 g SO3
Percentage yield = (actual yield / theoretical yield) x 100
Percentage yield = (12 g / 18.1 g) x 100 = 66.3% (approx)
2 SO2 + O2 -> 2 SO3
From the equation, we can see that 2 moles of SO2 react with 1 mole of O2 to produce 2 moles of SO3.
First, we need to calculate the limiting reagent to determine the theoretical yield.
The moles of SO2 present = 14.5 g / 64.06 g/mol = 0.226 moles
The moles of O2 present = 21 g / 32 g/mol = 0.656 moles
From the above values, we see that O2 is in excess as it has more moles than required for the reaction. Therefore, SO2 is the limiting reagent.
Theoretical yield = (0.226 mol SO2) x (2 mol SO3/2 mol SO2) x (80.06 g/mol SO3) = 18.1 g SO3
Percentage yield = (actual yield / theoretical yield) x 100
Percentage yield = (12 g / 18.1 g) x 100 = 66.3% (approx)