Question M.3

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Annah Khan 1B
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Joined: Fri Jul 22, 2016 3:00 am

Question M.3

Postby Annah Khan 1B » Sat Jul 01, 2017 7:14 pm

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?

I was able to convert the molar mass into moles, but I had trouble going from there. I wasn't sure why we start with CaCO3 instead of CO2 in order to find the limiting reactant.

Posts: 24
Joined: Fri Feb 24, 2017 3:03 am

Re: Question M.3

Postby 405060280 » Sat Jul 01, 2017 8:17 pm

Well, there is only one reactant in this problem, so I don't think there exists a limiting reactant. In this problem, since all you need to find is the percentage yield, you do not need to calculate the mole of the product (CO2, in this case). First of all, calculate the mole of CaCO3 based on it's mass, and then figure out how many moles of CO2 it will produce (the mole number will be the theoretical amount of product). Then, convert moles of CO2 into mass, which you already know how to do. Finally, use equation: percentage yield= (actual yield)/(theoretical yield) *100%. And that'll be your answer! Let me know if you have any questions.

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