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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.

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