## Question

Determine the percentage of calcium carbonate (CaCO3) in a rock sample. You crush up a 20g sample and dissolve it in excess dilutre hydrochloric acid. The evolved gas (CO2) is collected under water and at 25 degrees celcius and 100Kpa the volume is found to be 3.43Litres.

a) Calculate the number of moles of CO2 evolved.
b) Calculate the percentage of calcium barbonate present in the rock.

I think the formula turns out to be:
CaCO[size=9]3[/size] + 2 HCL --> CaCl[size=9]2[/size] + H[size=9]2[/size]O + CO[size=9]2[/size]

Plz help me. Working out would be most helpful.
Thx.

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### You can also use basic rock

You can also use basic [url=http://www.stone-mill.com/mobile_crusher.html]rock crushing[/url] equipment for this job. Then do the remainder of the experiment, i think that will make the work a little more precision level.

Given that at RTP, one mole of (ideal) gas occupies about 24dm? (litres) [you had better check the exact figure you are taught], you should be able to work it out.

So you can work out moles of CO[size=7]2[/size], and from that the number of moles of CaCO[size=7]3[/size], and from that the mass of CaCO[size=7]3[/size], so finally the % of CaCO[size=7]3[/size] in the rock.

I am sorry but i don't understand the RTP and dm3 part.

So the formula to find moles would be n=m/M
right? but i can't see the grams :?

:( i am hopeless with moles..

Auch come on, moles ain't so bad...

"Moles" is just simply counting how many "molecules" (or atoms or ions) there is in something.

I'm sure if I gave you a bag of marbles, you could count how
many marbles there were in it.

The concept of moles isn't really any different.

It's just like a "dozen".
A "dozen" is just 12 things.
A "mole" is just six hundred thousand million million million things! (6 x 10^23 )

[quote="allan_chemist"]So you can work out moles of CO2, and from that the number of moles of CaCO3, and from that the mass of CaCO3, so finally the % of CaCO3 in the rock.[/quote]

Can someone help me.
I don't see how u get it..

OK....

Firstly, for some reason, when God invented the universe,
he decided that if you have a mole of gas molecules
(that is, if you have six hundred thousand million million million gas molecules),
at "room temperature and pressure" ("RTP" - the temperature and pressure of God's living room is 20oC and 1atmosphere ;-),
then the size the gas will take up is ONE LITRE ( = 1 cubic decimetre, 1dm3)

It doesn't matter what type of gas molecules they are - carbon dioxide,
chlorine, or even 'Lynx' spray-on deodorant -
if you have a mole of them, they take up 1dm3 of volume in space (at RTP)
(this is a bit odd, cos you might expect that the same number of bigger molecules would take up more space than the same number of smaller molecules - after all, a dozen apples take up a lot more space than a dozen baked beans,
and a dozen cabbages take up a lot more space than a dozen peas.... however, that's the way it is for gasses!) (*Ideal* gasses, anyways ;-)

ANYWAY,
in your question, it says "The evolved gas (CO2) is collected under water and at 25 degrees celcius and 100Kpa the volume is found to be 3.43Litres"

25oC and 100 kiloPascals is more or less "RTP" (room temperature and pressure).
And carbon dioxide is most definately a gas.
And God fixed it so that if you had six hundred thousand million million million gas molecules (ie, 1 mole of gas molecules), they'd take up 1dm3 (1 litre)....
.....so if they take up 3.43 litres,
then you must have 3.43 moles of the gas.
(ie 18 hundred thousamd mill........ oh you get the idea ;-)

Next step -
look at the equation
CaCO3 + 2 HCL --> CaCl2 + H2O + CO2

It says that one mole of CO2 is produced by one mole of CaCO3

so if 3.43 moles of CO2 were produced, you must have started off with 3.43 moles of CaCO3, yes?

can you get the rest of it now?
- how much does 3.43 moles of CaCO3 weigh?
- and how much did the original rock weigh? (20g, the question said)
--- so you should be able to see what %age of that 20g was the 3.43 moles of CaCO3

All correct.
Only 1 mole occupies about 24 litres, not 1.

[PV = nRT, p=100000Pa, n=1, R=R, T=298.15 (25?C)]

Yes, but equation pV=nRT is correct in any conditions (in any T,p and V).

R=8.314 Pa m3/ K mol
m3=metre cubic

[quote="feline1"]- how much does 3.43 moles of CaCO3 weigh?[/quote]

mass of CaCO[size=9]3[/size] = 3.43 x 100(relative mass) = 343 grams?
but that wouldn't be right... 343/20g???

:shock:

thx for explanation though

[quote="<chem>"]Yes, but equation pV=nRT is correct in any conditions (in any T,p and V).

R=8.314 Pa m3/ K mol
m3=metre cubic[/quote]
I do not deny this. What I was pointing out is that at 25?C & 100kPa, 1 mole of an ideal gas will occupy approximately 24dm? (0.024788191 m?), and not 1 litre, as was suggested.

[b]Minny:[/b]
If you were to have 3.43 moles of CaCO[size=7]3[/size], you would have 343 grams. But you do not have 3.43 moles.

if:
24 litres = 1 mole then
1 litre = 1/24 moles
thus 3.43 litres = 3.43/24 moles = 0.143 moles (approximately)

so you actually have 14.3 grams of CaCO[size=7]3[/size].

I'll let you work out the %!

WebElements: the periodic table on the WWW [http://www.webelements.com/]