Lab calculation question?

Question

"a Tums tablet (active ingredient: CaCO3) was dissolved in 25mL 0.6M HCL. It required 6 mL 0.6M NaOH to reach the titration's equivalence point."

the balanced equation for the reaction is CaCO3 + 2HCl --> CaCl2 + H2CO3

Where I'm stuck:
-Calculate the millimoles of HCl that reacted with the CaCO3

-Determine the mass (in mg) of CaCO3 in the Tums tablet


Thanks for any help!

Answer 1

CaCO3 + 2HCl --> CaCl2 + H2CO3 ---> CaCl2 + H2O + CO2(gas)
Thus released CO2, instead of H2CO3, will not react with NaOH to consume NaOH.
Above reaction takes place in an acidic environment. The HCl left over would then be titrated with NaOH to the end point.
6 mL 0.6M NaOH is 0.0036 Mole of NaOH, which can react with 0.0036 Mole of HCl. We started with 25mL 0.6M HCL, or 0.015 Mole of HCl. Clearly 0.015 - 0.0036 = 0.0114 Mole of HCl was consumed when react with 0.0057 Mole of CaCO3. That is 100.087g/Mol*0.0057Mol = 0.57g.