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Textbook question M3

Posted: Sun Oct 07, 2018 3:54 pm
by mbaker4E
When limestone, which is principally CaCO3, is heated, carbon dioxide and quicklime, CaO, are produced by the reaction CaCO3(s) --> CaO(s) + CO2(g). If 17.5g of CO2 is produced from the thermal decomposition of 42.73g of CaCO3, what is the percentage yield of the reaction?

How would you go about finding the percentage yield from the reaction CaCO3(s) --> CaO(s) + CO2(g)?

Re: Textbook question M3  [ENDORSED]

Posted: Sun Oct 07, 2018 4:08 pm
by Edward Xie 2E
Since the question gives the actual yield of 17.5g of CO2, the theoretical (maximum) yield would have to be found in order to calculate the percentage yield.

Using stoichiometry, the given mass of CaCO3 is converterd into moles, which is then used to find how many moles of CO2 are produced from the reaction. After that, the resultant moles of CO2 is converted to its mass in grams:

42.73g CaCO3 * (1 mol CaCO3/100.1g CaCO3) * (1 mol CO2/1 mol CaCO3) * (44.01g CO2/1 mol CO2) = 18.79g CO2

Percentage yield = (actual yield/theoretical yield) * 100%

(17.5g/18.79g) * 100% = 93.1% -> percentage yield