stochiometry?

Question

A piece of chalk (mainly calcium carbonate) is placed in 250. mL of 0.302 M HCl.
All the CaCO3 reacts, releasing carbon dioxide gas, and leaving a clear solution.
30.00 mL of the solution is pipetted into another flask.
45.3 mL of 0.0592 M NaOH is required to titrate the HCl remaining in this 30.00-mL portion.
What was the original mass of CaCO3 in the piece of chalk?

Answer 1

We start out with:

0.302 M HCl * .03 L = .00906 mol HCl used in the 30mL solution

then we're told there was an excess of HCl so we have to find out what that excess is:

0.0592 M NaOH * 0.0453L NaOH * 1 mol HCl / 1 mol NaOH =

= .00268176 mol HCl excess

Now we find the total amount of HCl that reacted with the chalk in the 30 mL solution:

0.00906 mol HCl - 0.00268176 mol HCl excess = 0.006378 mol HCl that reacted with the chalk in the 30 mL solution

Now we find the concentration of HCl that was used in the reaction by:

0.006378 mol HCl / .03 L = 0.212608 M HCl

This is the concentration of HCl that was used to react with the chalk so:

0.212608 M HCl * 0.25L (the total volume) = 0.053152 mol HCl that reacted with the chalk

Now we take this total mols of HCl that reacted with the chalk and setup stoic to find the mass of calcium carbonate:

0.053152 mol HCl * 1mol CaCO3/ 2mol HCl * 100.09 g CaCO3 / 1 mol CaCO3 =

=2.65999 g of CaCO3 in the piece of chalk

Answer 2

The NaOH will neutralise all of the HCl which didn't react with the chalk. So start by working out the moles of NaOH, then the moles of HCl in the 30ml. Scale that up to 250ml, and halve the number of moles, because every mole of CaCO3 reacts with 2 moles of HCl.
Then multiply by 100.

Answer 3

Molarity of soln. after CaCO3 has reacted = 45.3*0.0592/30 = 0.0894 M of HCl

So, moles of HCl consumed = 250*0.302 - 250*0.0894 = 53.152 millimoles

So, CaCO3 mass = (53.152*100*10^-3)/2 g = 2.66 g