calcium carbonate q (1 Viewer)

Libbster

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

richz

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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

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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

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xrtzx said:
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

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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
 
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rama_v

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xrtzx said:
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

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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 :D
 

Hardenne

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

richz

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rama_v said:
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 :rolleyes:

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

thnx :)
 
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katietheseal

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xrtzx said:
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 :rolleyes:

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:D
 

o0_jolie_0o

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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:D

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

falltopieces

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xrtzx said:
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

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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
 
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