$15.0~\mathrm{mL}$ of $1.4~\mathrm{M}\ \ce{HCl}$ was mixed with $1.00~\mathrm{g}$ of limestone (impure $\ce{CaCO3}$) until all the solid had dissolved. The solution was then transferred to a conical flask and made up to $200~\mathrm{mL}$ with water. A $20.0~\mathrm{mL}$ portion was then neutralised by $8.50~\mathrm{mL}$ of a $0.1~\mathrm{M}\ \ce{NaOH}$ solution.
Calculate:
- Amount of excess $\ce{HCl}$ in the $20.0~\mathrm{mL}$ portion
- Amount of excess $\ce{HCl}$ in the $200~\mathrm{mL}$ portion
- Amount of $\ce{HCl}$ which reacted with $\ce{CaCO3}$
My attempt:
When the limestone is dissolved,
$$\ce{2 HCl(aq) + CaCO3(s) -> CaCl2(aq) + CO2(g) + H2O(l)}$$
$$n(\ce{HCl}) = c \times V = (15 \times 10^{-3}~\mathrm{L}) (1.4~\mathrm{M}) = 0.021~\mathrm{mol}$$ $$n(\ce{CaCO3}) = x~\mathrm{mol}$$
When the solution is diluted to $200~\mathrm{mL}$ the molarity of $\ce{HCl}$ can be found by, \begin{align} n_\mathrm i &= n_\mathrm f\\ c_\mathrm i V_\mathrm i &= c_\mathrm f V_\mathrm f\\ c_\mathrm f &= \frac{c_\mathrm i V_\mathrm i}{V_\mathrm f}\\ \therefore c_\mathrm f (\ce{HCl}) &= \frac{(15\times 10^{-3}~\mathrm{L})(1.4~\mathrm{M})}{(200\times 10^{-3}~\mathrm{L})} = 0.105~\mathrm{M} \end{align}
I'm having doubts about my working up until this stage. Am I correct so far?