Posted by Jordan on Tuesday, December 2, 2008 at 10:22pm.
Check the numbers in your post. Either I'm on the fast track to nowhere or there is at least one wrong number. I'm getting 150+% purity and that can't be.
Those are all the numbers in the problem. I worked further in the problem and found...
(a) 15.9/20.29= 78.54%
and for part (b)
I found that 20.29g of MgCO3 makes .24mol CO2. So....
Pressure of CO2 = 12.54 atm.
Does that seem right?
yes and no.
MgCO3 + heat ==> MgO + CO2
so the 15.9 g of the residue is not MgCO3, it is MgO and if the 20.29 g MgCO3 sample at the beginning was 100% pure it can produce ONLY 9.69 g MgO (check it out. 20.29/molar mass MgCO3 about 0.24 mols (what you had) and that times the molar mass of MgO is 9.69 g. So you see the problem I'm having. Or, if we work backwards and have 15.9 g MgO at the end of the decomposition,
15.9/molar mass MgO give mols MgO and that's the same as mols MgCO3 so that times molar mass MgCO3 gives 33. something grams MgCO3 and you didn't have but 20.29 to begin with.
As far as the pressure of CO2, you can't use 10.59 g CO2 because that's how much you would obtain IF the MgCO3 was 100% pure and it isn't supposed to be 100%. But yes, you use PV = nRT.
Given the data, I would use the 15.9 g MgO, determine mols of that and grams CO2 from that, then PV = nRT but that works only if the 15.9 is a good number.
So that would give you .39 mol. Which would then give you P = 20.77 atm.
I see what you are saying about the numbers in this problem. Unfortunately my teachers often gives problems that seem to have something missing, and I don't know what to do about it.
I was also wondering if you could explain to me why you said I should use the 15.9g MgO to find mol CO2. That is what I originally thought to do, but then I spoke to a tutor who is the one that said to use 20.29g to find moles CO2.
Thank you again.