You want to measure the CaCo3 contents in chalk. To this end, you grind a piece of chalk, dissolve the slurry to make 10ml of solution and drive the carbonate out by adding 10 ml of 2M HCl. You titrate the remaining HCl back with 0.1M potassium hydroxide solution, of which you need 47 ml. Assume all carbonate was CaCO3, how much was it in grams?

moles HCl added initially = M x L = ??

Some was not used (there was an excess of HCl); how much of an excess did you have? That will be moles KOH = M x L = ??

So the difference between moles initially and moles of excess must be the amount CaCO3 in the chalk. Then moles x molar mass CaCO3 = g CaCO3.
Post your work if you get stuck.

How would you calculate the excess of HCl?

Go back and re-read my response. The excess HCl was titrated with 47 mL of 0.1 M KOH. M x L = moles KOH added which is the same as moles HCl in excess.

You need to make a correction, then, for the fact that CaCO3 used two moles HCl for every one mole CaCO3.
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2

So initial moles HCl - moles HCl in excess = moles HCl used to neutralize the CaCO3.
Then 1/2 moles HCl used = moles CaCO3. Then g CaCO3 = moles CaCO3 x molar mass CaCO3

How do you find the initial moles of HCl?

To find out the amount of CaCO3 in grams, we need to determine the number of moles of HCl that reacted with CaCO3 during the titration. From the given information, we know that 10 mL of 2M HCl reacted completely with the carbonate in the chalk solution.

First, let's calculate the number of moles of HCl used:

Moles of HCl = Volume of HCl (in L) × Concentration of HCl (in mol/L)

Given that we used 10 mL of 2M HCl, we can convert the volume to liters:

Volume of HCl = 10 mL = 10/1000 L = 0.01 L

Now, let's calculate the moles of HCl used:

Moles of HCl = 0.01 L × 2 mol/L = 0.02 mol

Since the balanced chemical equation between HCl and CaCO3 is 2HCl + CaCO3 -> CaCl2 + H2O + CO2, we see that 2 moles of HCl react with 1 mole of CaCO3.

Therefore, the moles of CaCO3 in the chalk solution are:

Moles of CaCO3 = 0.02 mol HCl × 1 mol CaCO3 / 2 mol HCl = 0.01 mol CaCO3

To find the mass of CaCO3, we need to use the molar mass of CaCO3, which is approximately 100.09 g/mol.

Mass of CaCO3 = Moles of CaCO3 × Molar mass of CaCO3

Mass of CaCO3 = 0.01 mol × 100.09 g/mol = 1.0009 g

Therefore, the amount of CaCO3 in grams in the chalk is approximately 1.0009 grams.