calcium carbonate q (1 Viewer)

Bokky

Member
ah crap! i got 1.4g now i remember that i timesed the moles of the HCL by 2 coz there was 2 moles of it. I shouldntve!!! grrr. Ah well, i recon i nailed the rest of the exam, the majority of it anyway

richz

Active Member
falltopieces said:
I got 0.75g too
unfortunately, we are rong , shud have read the q. any 06ers out there, my lesson to u is to read qs

Xenocide

Member
You work out that there are 0.15 moles of HCl:

You work out how many moles of NaOH are used, something like 0.0138 or something. Then u realise that the amoutn of carbonate must have neutralised (0.15-0.0138) moles of HCL. You then halve this due to equation and multiple by molar weight. Correct answer should be around 0.6x grams.

~ ReNcH ~

!<-- ?(°«°)? -->!
katietheseal said:
im pretty sure you had to take the excess number moles of hcl away from the original which was 0.015-something like 0.0014, so 0.0136 moles of hcl reacted with the CaCO3, then mass = 0.0136x100.09 or something like that, so about 1.36 grams i got. but everyone getting o.6 has got me worried

lol, but chems over
The molar ratio was such that you had to halve the number of moles i.e. 0.5 x 0.0136 x 100.09 = 0.68g.

tabularasa

Member
i got hcl to have 24 mols whihc was heaps weird. i don;t htink i got this one right at all.

katietheseal

New Member
~ ReNcH ~ said:
The molar ratio was such that you had to halve the number of moles i.e. 0.5 x 0.0136 x 100.09 = 0.68g.
damn, yep u right im pretty sure, i probably, stupily messed up the simple formula, hope they're not too harsh in the marking

serge

Member
~ ReNcH ~ said:
The molar ratio was such that you had to halve the number of moles i.e. 0.5 x 0.0136 x 100.09 = 0.68g.
that was for the carbonate but the NaOH was 1:1

Numero Uno

Active Member
~ ReNcH ~ said:
The molar ratio was such that you had to halve the number of moles i.e. 0.5 x 0.0136 x 100.09 = 0.68g.
hmm yeah it does seem right
i got that as well
but for some reason i times it by two for some reason ..
well there goes a mark .. oh well

serge

Member
so im thinking, 1/2 whatever u got as a HCL number of moles...

then minus the number of NaOH moles used up in neutralisation
and then times that whole result by the molar mass 100.9 or something..

jynxe

Member
I think NaOH, from memory was like 0.014 L and .1 molar so it was 0.0014 moles.
That's the amount required to neutralise the extra HCl, so the HSC used in the original reaction is 0.015 - 0.0014 = 0.0136 moles.

But, from the reaction, the molar ratio of HCl: CaCO3 is 2:1

so the moles of CaCO3 = 6.8 x 10^-3
and thus mass (m = n.M) = 6.8 x 10^-3 x (40.08+ 16 x 3 + 12.01) = 0.680612 grams.

I hope =P

PS hey Serge it's Erica =P

Last edited by a moderator:

~ ReNcH ~

!<-- ?(°«°)? -->!
serge said:
that was for the carbonate but the NaOH was 1:1
Yea, the ratio required in the titration part was 1:1 i.e. HCl:NaOH. But in the second part (to find the mass of calcium carbonate), you had to use the first equation...I think it was part a) and in this equation the molar ratio was 2 HCl: 1 CaCO3

souha882002

New Member
b_flat_major THATS EXACTLY WHAT I DID! And i got the same answer as you, but i was worried about rounding off because last year they were marked down for incorrect significant figures . i just put it to 2 sig figures.........i hope we did it the right way!

AreYouAlright?

Actuarial Co-op 2006
Dam. I knew it was something like that but i was subtracting 0.015 from the value we get from the titration and it was negative..... *hits head on desk* should have known the subtraction was other way around...

Dumsum

has a large Member;
FUCK I think I missed a decimal place in one of my calculations... so I was out by like a factor of 10. There goes a mark...

banana_monkey

Better than you
~ ReNcH ~ said:
The molar ratio was such that you had to halve the number of moles i.e. 0.5 x 0.0136 x 100.09 = 0.68g.
Yeah, that sounds right. That was my reasonining at least. I'm pretty sure I also got around 0.6something. I found the question a bit tough though, I've never ran into a question were you had to take into account the excess and use it to find the concentration of the original substance. Never seen it in ANY book I had. Damm you BOS!

alluring

New Member
ahhh fck meee,
fckn dont know why i forgot to put water as a product in the reaction
fckkk
how many marks would i lose? does anyone know?

serge

Member
alluring said:
ahhh fck meee,
fckn dont know why i forgot to put water as a product in the reaction
fckkk
how many marks would i lose? does anyone know?
yes, BOS chem markers know

serge

Member
b_flat_major said:
PS hey Serge it's Erica =P
hey, howd u find multiple choice, 15 no doubt?

Haku

Member
15 is A

but for the titration i got 0.25g...

jynxe

Member
serge said:
hey, howd u find multiple choice
I haven't checked! I didn't bring my exam paper out with me (I keep leaving it on my desk) but I hope I did alright with it. I bet you did well. Just one more to go. Physics and then woooooo