1. May 10, 2012

thepoll

Hi, I've been trying and trying but I can't seem to solve this. Hopefully, there's someone here kind enough to help me through :X

2.5g of impure CaCO3 was dissolved in 25.0cm^3 of 2.00 mol dm^-3 HCL in a 100cm^3 volumetric flask and made up to the mark with distilled water. This solution is labelled as Solution X. 15.0cm^3 of this mixture required 20.0cm^3 of 0.0500 mol dm^-3 NaOH for neutralization.

Calculate the number of moles of HCL that has reacted with 2.5g of CaCO3.

1. I calculated the mole of HCL in 25.00cm^3 of 2.00 mol dm^-3 solution and also the number of moles of NaOH required to be titrated with 25cm^3 of the mixture of X.
2. I then tried calculating the concentration of Solution X that contains excess HCL. However, I got stuck there. If anyone could just hint me the next step, I would greatly appreciate it! thanks:)

2. May 10, 2012

Staff: Mentor

How many moles of HCl initially?

How many moles of NaOH reacted with the excess HCl?

How much of the excess HCl was present?

Then it is just initial minus excess.

3. May 11, 2012

thepoll

thanks! I managed to solve it. a lil tricky though (: