titration... moley moles...

Can you write out the balanced chemical equation?

CaCO3 + 2HCL ---> H20 + CO2 + CaCl2

??

Yup

So, you're told that the sample reacted with 100cm³ of 0.200 mol dm^-3 HCl, work out how many moles you have of this.

Next step is to work out the moles of NaOH (this tells you the amount of acid left over AFTER the reaction with the carbonate).

mol of HCL = 0.02
mol of NaOh = 0.00248

So how many moles of acid reacted with the carbonate in the limestone sample?

Once, you've got this number you can work out how much carbonate was present in the sample

Next step is to work out the moles of NaOH (this tells you the amount of acid left over AFTER the reaction with the carbonate).

I thought that the number of moles for the carbonate is simpply 0.01 mol (half mole of HCL). I dont understand when and where to use the NaOH mol for? :/

The limestone was added to a known amount of acid (which you've worked out). After the reaction the solution was still acidic i.e. there was more acid than was needed to neutralise the sample.

The NaOH was used to work out how much acid exactly was left over. It reacted by neutralisation (HCl + NaOH ---> NaCl + H₂O). So...

[amount of acid used in reaction] = [amount at start] - [amount at end]

then you convert the amount of acid to carbonate using the equation I asked you for at the start

p.s. it's HCl not HCL

how did you get 0.02?

A 1.00g sample of limestone is allowed to react with 100cm3 of 0.200 moldm^-3 HCL. The excess acid required 24.8 cm^3 of 0.1 mol dm^3 NaOH sol. Calculate the percentage of calcium carbonate in the limestone.

Anyone?

2HCl + CaCO3 --> CaCl2 + CO2 + H2O
HCl + NaOH --> NaCl + H2O
- mol NaOH = 24.8/1000 x 0.1 = 0.00248
- mol HCl = 100/1000 x 0.2 = 0.02
- XS acid moles = 0.02 - 0.00248 = 0.01752
- mole ratio between CaCO3 and HCl is 1:2 therefore divide 0.01752 by 2 which gives 8.76 x 10-3
- mol CaCO3 = 0.00876
- mass CaCO3 = moles x Mr = 0.00876 x (40.1 + 12 + 48) = 0.877g
- % = 87.7%

2HCl + CaCO3 --> CaCl2 + CO2 + H2O
HCl + NaOH --> NaCl + H2O
- mol NaOH = 24.8/1000 x 0.1 = 0.00248
- mol HCl = 100/1000 x 0.2 = 0.02
- XS acid moles = 0.02 - 0.00248 = 0.01752
- mole ratio between CaCO3 and HCl is 1:2 therefore divide 0.01752 by 2 which gives 8.76 x 10-3
- mol CaCO3 = 0.00876
- mass CaCO3 = moles x Mr = 0.00876 x (40.1 + 12 + 48) = 0.877g
- % = 87.7%