calcium carbonate q (1 Viewer)

[Libbster]

what did everyone get for the weight of calcium carbonate in that q where there was the titration etc etc?

 

[n00b]

something

it was like 0.6g wasn't it

 

[Abtari]

yeah, i think it was 0.6 g

 

[-Swifty-]

i got something small less and 1.0 g

 

[queenie]

yea, something like that

 

[rama_v]

I got 0.65 g

 

[dangerousdave]

If this is the question where they added HCl then NaOH, then i got 0.9996... so i rounded up to 1g.
I get the feeling i'm very wrong.

 

[richz]

well, the eqn was CaCO3 + 2HCl -> H2O + CO2 + CaCl2

and the concentration of HCl, was 0.6*0.025= 0.015moles
since HCl 2* CaCO3 so 0.0075moles of CaCO3
moles = g/molar mass
g=0.0075*100= 0.75g

correct me if im rong

 

[jasont88]

I can't even remember what I got.....think 0.6 does ring a bell though....I can't be too sure on that.....I aint holding my breath on my calculation questions...but I kicked arse in the 4,5 and 7 mark questions!!!

 

[rama_v]

well, the eqn was CaCO3 + 2HCl -> H2O + CO2 + CaCl2

and the concentration of HCl, was 0.6*0.025= 0.015moles
since HCl 2* CaCO3 so 0.0075moles of CaCO3
moles = g/molar mass
g=0.0075*100= 0.75g

correct me if im rong

You didnt include the titration stuff
That actually reduces the moles of HCL available

 

[richz]

ok... dang i shudve read the q. anyway, i dont remeber the rest of the q. do u remeber?? can u show me ur working

 

[rama_v]

so i dont remeber the rest of the q. do u remeber?? can u show me ur working

Umm, sorry no offence m8 I didn't mean to pick on ya
I can't remember it, but it was a 1:1 ratio so you had to subtract the moles of Hydrogen from this reaction or somethign..

 

[justchillin]

U wrote an equation: 2HCl + CaCO3 ...
Moles of HCl worked out using concentration times volume (litres)
Then the titration worked out excess of HCl, so u found number of moles HCL excess using voulme and conco of NaOH then mole ratios. Then u subtracted the excess from the original mole of HCl, used mole ration again to get mole of CaCO3...then mass CaCO3, 0.6 or smthing was correct

 

[Hardenne]

I think you had to take into consideration the excess HCl, that wasnt used up in the reaction?

 

[richz]

Umm, sorry no offence m8 I didn't mean to pick on ya
I can't remember it, but it was a 1:1 ratio so you had to subtract the moles of Hydrogen from this reaction or somethign..

no, no its fine, im just posting up my thoughts, i even said in that post "correct me if im rong." Maybe i should have read the q properly

if anyone remebers the rest of the q, can u finish the answer

thnx

 

[katietheseal]

no, no its fine, im just posting up my thoughts, i even said in that post "correct me if im rong." Maybe i should have read the q properly

if anyone remebers the rest of the q, can u finish the answer

thnx

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

 

[o0_jolie_0o]

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

hmmz.. yeh I got that to... hmmz... hopefully they can read my working out.. I mean its fully scribbled >_<

 

[Luthien]

i got .65 or .67 or something like that

 

[falltopieces]

well, the eqn was CaCO3 + 2HCl -> H2O + CO2 + CaCl2

and the concentration of HCl, was 0.6*0.025= 0.015moles
since HCl 2* CaCO3 so 0.0075moles of CaCO3
moles = g/molar mass
g=0.0075*100= 0.75g

correct me if im rong

I got 0.75g too

 

[serge]

i got .608 grams

by halving the number of moles of HCl and minusing the
number of moles of NaOH used from that HCl value