# Application of limiting reagents and acid-base stoichiometry

I'm having a bit of trouble with the latter questions of this problem.

$$\pu{15.0 mL}$$ of $$\pu{1.4 M}$$ $$\ce{HCl}$$ was mixed with $$\pu{1.00 g}$$ of $$\ce{CaCO3}$$ until all the solid had dissolved. The solution was then transferred to a conical flask and made up to $$\pu{200 mL}$$ with water. A $$\pu{20.0 mL}$$ portion of this solution was then neutralized by $$\pu{8.50 mL}$$ of a $$\pu{0.100 M}$$ $$\ce{NaOH}$$ solution.
Calculate:

1. Amount of substance in excess of $$\ce{HCl}$$ in the $$\pu{20.0 mL}$$ portion
2. Amount of substance in excess of $$\ce{HCl}$$ in the $$\pu{200.0 mL}$$ portion
3. Amount of substance in excess of $$\ce{HCl}$$ which reacted with $$\ce{CaCO3}$$

My approach

$$\ce{2HCl(aq) + CaCO3(s) -> CaCl2(aq) + H2O(l) + CO2(g)}$$

Finding the amount of substance of each reactant and hence the limiting reactant,

For $$\ce{HCl}$$: \begin{align} M[\ce{HCl}] &= \frac{\mathrm{mol}}{\mathrm{L}}\\ \mathrm{mol}[\ce{HCl}] &= \mathrm{M}\times \mathrm{L}\\ \therefore \mathrm{mol}[\ce{HCl}] &= 1.4\times (15.0\times 10^{-3}) = 0.021 \end{align}

For $$\ce{CaCO_3}$$: \begin{align} \mathrm{mol}[\ce{CaCO_3}] &= \frac{\mathrm{m}}{\mathrm{mm}}\\ \mathrm{mol}[\ce{CaCO_3}] &= \frac{1.00}{100.0869 }\\ \therefore \mathrm{mol}[\ce{CaCO_3}] &= 9.9913\times10^{-3} \end{align}

Therefore (after dividing by stoichiometric coefficients) $$\ce{CaCO_3}$$ is limiting.

From here I'm not sure if what I did next is correct. I found the amount of substance of $$\ce{CaCl_2}$$ produced the remaining $$\ce{HCl}$$ in excess. Which were,

\begin{align} \mathrm{mol}[\ce{CaCl_2}] &= 9.9913\times10^{-3} \text{ and,}\\ \mathrm{mol}[\ce{HCl}] &= 1.1009\times10^{-2} \end{align}

(Meaning that we have $$9.9913\times10^{-3}\mathrm{mol}$$ of $$\ce{CaCl_2}$$ and $$1.1009\times10^{-2}\mathrm{mol}$$ of $$\ce{HCl}$$ in in $$\pu{15 mL}$$ of solution?)

From here on I'm stumped.