Question #202653

A 11.3 sample of CaCO_{3} was treated with aqueous H_{2}SO_{4}, producing calcium sulfate, water and 3.65 g of CO_{2}(g). What was the % yield of CO_{2}?

Expert's answer

CaCO3 mass is 11.3g

Then the number of moles will be given by weight divided by molecular mass

No of mole=weight/molecular weight

=11.3g/100=0.113moles

The balanced chemical equation is CaCO3(s) + H2SO4 (aq) => CaSO4(s) + H2O + CO2 (g)

From this equation one mole of CO2 will be produced by one mole of CaCO3. Therefore 0.113moles of CaCO3 will produce 0.113 moles of CO2

The weight of 0.113 moles of CO2= 0.113moles*44=4.972g

But it produces 3.65g of CO2

The percentage yield will be 3.65g/4.972g=73.41%

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