# Chemistry

I'm working on a problem that says: A 4.00 gram sample of a mixture of CaO and BaO is placed in a 1.00 L vessel containing Co2 at a pressure of 730 torr and a temperature of 25C. The CO2 reacts with the CaO and BaO forming CaCO3 and BaCO3. When the reaction is complete, the pressure of the remaining Co2 is 150 torr.

a) Calculate the number of moles of Co2 that have reacted.

I already did this part and got 0.0312 moles CO2. I am stuck on the second part though.

b) Calculate the mass percentage of CaO in the mixture.

I know it will have something to do with grams of CaO = x and grams BaO = 4-x but I don't know how to solve from there. If anyone could please help that would be great thanks!

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1. You started right; just stopped too soon. The equations are
CaO + CO2 ==> CaCO3
BaO + CO2 ==> BaCO3

If x = g CaO
and 4-x = g BaO, then
mols CO2 reacting with CaO = x/molar mass CaO.
mols CO2 reacting with BaO = (4-x)/molar mass BaO
Then mols CO2 reacting with CaO + mols CO2 reacting with BaO = 0.0312
Solve for x and after that
%CaO = (g CaO/mass sample)*100 = ?
%BaO = (g BaO/mass sample)*100 = ?
Post your work if you get stuck.

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2. Hi I did what you said and got 11.32% for CaO, would that be right? Thank you!!

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3. Yes. Good work. Let me show you how to check that number. I almost always check my numbers some third way to see if I get the same answer.
If CaO = 11.32%, then BaO must be 100-11.32 = 88.68%. So in the 4 g sample, we have
g CaO = 0.1132 x 4 = 0.4528 g
g BaO = 0.8868 x 4 = 3.547 g

mols CaO = 0.4528/56.07 = 0.008076
mols BaO = 3.547/153.3 = 0.02314

mols CO2 used by CaO = 0.008076
mols CO2 used by BaO = 0.02314
total mols CO2 used = 0.03122. I used too many places (too many significant figures) but you see this 0.0312 mols CO2 agrees with what we started with. If that 11.32% had been wrong, I wouldn't have ended up with 0.0312 mols CO2 used by the BaO + CaO. And this may be a good time to point out that if your problem stated that the start was with 4 g (and not 4.0 or 4.00 etc), then you are allowed only 1 place in the final answer.

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