Question

A 0.450-gram sample of impure CaCO3(s) is dissolved in 50.0 mL of 0.150 M HCl(aq). The equation for the reaction is:
CaCO3(s)+2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g)

The excess HCl(aq) is titrated by 9.75 mL of 0.125 M NaOH(aq). Calculate the mass percentage of CaCO3(s) in the sample.



Here's what I did. Is this correct? If not how do I approach it?:
excess HCl - x/.00975 L = 0.125 M
x = .00121875 mol

total HCl - x/0.05 L = 0.150 M
x = 0.0075 mol

0.0075 - 0.00121875 mol = 0.00628125 mol
0.00628125 mol HCl X 1mol CaCO3/2mol HCl X 100.0869 g/1 mol CaCO3 = 0.3143 g
0.3143 g/0.450 g = 69.8%?

Answer 1

That looks good to me.