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Percentage Yield M.3

Posted: Tue Oct 03, 2017 12:19 am
by Andrew Nguyen 2I
So for this problem, it's saying that from the calcium carbonate's decomposition, a certain amount of CO2 is produced. I know that the percentage yield should be the actual yield over the theoretical yield.

But just to confirm, do I assume that the theoretical yield in this case is 42.73 g (100% for this situation?) ; and it has something to do with the mole ratios correct ?

Re: Percentage Yield M.3

Posted: Tue Oct 03, 2017 9:10 am
by Naveed Zaman 1C
You are partially right. What you have to realize is that they don't actually give you all the information you need. Remember that BOTH the theoretical yield and the actual yield have to be for the same thing (in this case, for CO2). What you have to do with the 42.73g of CaCO3 is use stoichiometry to convert to grams of CO2; this becomes your theoretical yield (you expect to form "x" grams of CO2 from 42.73g of CaCO3). You are already given the actual yield: 17.5g. All that's left is plugging these values in the %yield formula to get the answer. Hope this helps!

Re: Percentage Yield M.3

Posted: Wed Oct 04, 2017 10:37 pm
by Amanda Hagen 1L
I just want clarify the stoichiometric calculations needed to get from 42.73 g of CaCO3 to the grams CO2 in case more explanation was needed:

1. Divide by the given grams of CaCO3 (42.73 g CaCO3) by the molar mass of CaCO3 (100.09 g.mol-1 CaCO3) to get moles of CaCO3.
2. Convert moles of CaCO3 to moles of CO2 by using a mole ratio (they will be the same number because it is a 1:1 ratio)
3. Convert moles CO2 to grams CO2 by multiplying by the molar mass of CO2 (44.01 g.mol-1 CO2).

Now this gives you the theoretical yield of CO2 and you can use it to find the percent yield as described above.