# Need someone to confirm my answer to Chem/Math problem.

• Oct 9th 2010, 10:22 PM
butters
Need someone to confirm my answer to Chem/Math problem.
I don't have the answer key, so I need someone to check and tell me if I got it right. I probably got it wrong, but there is a slim chance I might have gotten this one right. And please check if I got the rounding off of the numbers right because I am sort of unsure of the rules (chemistry rules in rounding off digits).

Problem is:
"What mass of lime CaO can be obtained by heating 125 kg of limestone that is 95.0% by mass CaCO3"

"CaCO3 = CaO = CO2"

- Note: the numbers 3 and 2 are subscript and indicate the amount of Oxygen.

The answer I get is 70100g/mol or 7.01x10^4
• Oct 9th 2010, 11:12 PM
Pim
If you'd like people to take a look if your calculation is correct, but you don't post the calculation, there is absolutely no way we can check it.

Secondly, 70100g/mole would mean 1 mole weighs over 70 kilogram. That is in no way possible. Hence the need to post the steps you took to get there.
• Oct 9th 2010, 11:18 PM
butters
Hold on...
http://img836.imageshack.us/img836/2392/moto0182.jpg

Is this any good?
• Oct 9th 2010, 11:24 PM
Jhevon
Quote:

Originally Posted by butters
I don't have the answer key, so I need someone to check and tell me if I got it right. I probably got it wrong, but there is a slim chance I might have gotten this one right. And please check if I got the rounding off of the numbers right because I am sort of unsure of the rules (chemistry rules in rounding off digits).

Problem is:
"What mass of lime CaO can be obtained by heating 125 kg of limestone that is 95.0% by mass CaCO3"

first note that you are working with 118.75 kg = 118750 g of \$\displaystyle \displaystyle CaCO_3\$ (95% of the limestone sample)
Quote:

"CaCO3 = CaO = CO2"
huh??

Quote:

- Note: the numbers 3 and 2 are subscript and indicate the amount of Oxygen.

The answer I get is 70100g/mol or 7.01x10^4
first, they didn't ask you for molar mass...secondly, i got 66 500 g CaO can be made. The Ca is the limiting reagent.
• Oct 9th 2010, 11:28 PM
Jhevon
Quote:

Originally Posted by butters
Hold on...
http://img836.imageshack.us/img836/2392/moto0182.jpg

Is this any good?

Nope. first note that the moles in your calculations were canceled. how is it they ended up in your answer? and remember, you don't have 125 kg of the relevant sample. only 95% of it is usable

other than that, your process was correct
• Oct 9th 2010, 11:33 PM
linalg123
you have not factored in that only 95% of the mass of the limestone is CaCO3. ie. 118.75kg CaCO3

firstly, work out how many mol CaCO3 you have using n(CaCO3)=m(CaCO3)/M(CaCO3)

then using n(CaCO3) = n(CaO) = m(CaO)/M(CaO)

rearrange to give m(CaO) = n(CaO) * M(CaO)

Does that make sense?
• Oct 10th 2010, 09:24 AM
butters
Quote:

Originally Posted by Jhevon
huh??

Sorry, CaCO3 is actually the reactant, that is heated, while CO2 and CaO are the products.

The book I am using does not show how to solve a problem such as this, but I think I got this.
Is this right, it's very close to the answer you guys are getting...

http://img156.imageshack.us/img156/8079/moto0183.jpg

It's a bit messy, but keep in mind that I just woke up, and I'm not really a morning person.

Okay, as you can see that I'm a bit off with the rounding off of the digits; if I round off the CaCO4 kg to three digits (because of the 125g value) then I get a result of 66679. Is that the right method?
• Oct 10th 2010, 09:32 AM
Jhevon
Quote:

Originally Posted by butters
Sorry, CaCO3 is actually the reactant, that is heated, while CO2 and CaO are the products.

The book I am using does not show how to solve a problem such as this, but I think I got this.
Is this right, it's very close to the answer you guys are getting...

http://img156.imageshack.us/img156/8079/moto0183.jpg

It's a bit messy, but keep in mind that I just woke up, and I'm not really a morning person.

Okay, as you can see that I'm a bit off with the rounding off of the digits; if I round off the CaCO4 kg to three digits (because of the 125g value) then I get a result of 66679. Is that the right method?

Not a morning person either. the morning is just over though.

anyway, yes, you are correct. I did more rounding than you did. I used whole numbers for the molar masses. now you just need to work on not making such a mess when you put your answer down on paper (Rofl)
• Oct 10th 2010, 09:41 AM
butters
Quote:

Originally Posted by Jhevon
Not a morning person either. the morning is just over though.

anyway, yes, you are correct. I did more rounding than you did. I used whole numbers for the molar masses. now you just need to work on not making such a mess when you put your answer down on paper (Rofl)

Mornings for me are in the 12:00 - 1:00 range (pm)... LOL
Anyway, so you are saying that my rounding off that led to 66679 is correct (there are some silly rules for this in Chem), or did you mean the one I did on the paper?
• Oct 10th 2010, 09:44 AM
Jhevon
Quote:

Originally Posted by butters
Mornings for me are in the 12:00 - 1:00 range (pm)... LOL
Anyway, so you are saying that my rounding off that led to 66679 is correct (there are some silly rules for this in Chem), or did you mean the one I did on the paper?

well, that wasn't the answer you got... but anyway, i am saying the discrepancy between your answer and mine is that i rounded more than you did. i don't think it would be a big deal here. your answer is fine and would probably be the more accepted one.

in general though, you want your answer to use and maintain the least amount of significant figures given in the problem. your answer is fine, don't worry about it.