[VanKwisH]

A piece of chalk (mainly calcium carbonate) is placed in 250. mL of 0.348 M HCl.
All the CaCO3 reacts, releasing carbon dioxide gas, and leaving a clear solution.
35.00 mL of the solution is pipetted into another flask.
28.8 mL of 0.0509 M NaOH is required to titrate the HCl remaining in this 35.00-mL portion.
What was the original mass of CaCO3 in the piece of chalk?

Well i first started with making the chemical formula

CaCO3 + HCl --> CaCl + CO2 + OH

Using m1v1 = m2v2
m1 * 0.035 = 0.0509 * 0.028.8
m1 = 0.04188 m/L of HCl
Now is where i got lost, can anyone help me get started with this?

[Borek]

http://www.titrations.info/back-titration