Sục từ từ CO2 vào V lít...

C

- Hướng dẫn giải

Đáp án C

$\begin{array}{l}+ \left\{\begin{array}{l}\mathrm{TN}1: {\mathrm{n}}_{{\mathrm{CO}}_2}=\mathrm{b} \mathrm{thì} {\mathrm{n}}_{{\mathrm{CaCO}}_3}=0,06\\ \mathrm{TN}2: {\mathrm{n}}_{{\mathrm{CO}}_2}=2\mathrm{b} \mathrm{thì} {\mathrm{n}}_{{\mathrm{CaCO}}_3}=0,08\end{array}\right.\\ \Rightarrow \left\{\begin{array}{l}\mathrm{TN}1: \mathrm{Ca}(\mathrm{OH}{)}_2 \mathrm{chưa} \mathrm{phản} \mathrm{ứng} \mathrm{hết}\\ \mathrm{TN}2: \mathrm{Ca}(\mathrm{OH}{)}_2 \mathrm{phản} \mathrm{ứng} \mathrm{hết}, {\mathrm{CaCO}}_3 \mathrm{bị} \tan \mathrm{một} \mathrm{phần}\end{array}\right.\\ \Rightarrow \left\{\begin{array}{l}\mathrm{TN}1: {\mathrm{n}}_{{\mathrm{CO}}_2}={\mathrm{n}}_{{\mathrm{CaCO}}_3}=0,06\\ \mathrm{TN}2: \underset{0,08}{\underbrace{{\mathrm{n}}_{{{\mathrm{CO}}_3}^{2-}}}}=\underset{?}{\underbrace{{\mathrm{n}}_{{\mathrm{OH}}^{-}}}}-\underset{0,12}{\underbrace{{\mathrm{n}}_{{\mathrm{CO}}_2}}}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{\mathrm{n}}_{{\mathrm{OH}}^{-}}=0,2\\ {\mathrm{n}}_{\mathrm{Ca}{\left(\mathrm{OH}\right)}_2}=0,1\end{array}\right.\Rightarrow {\mathrm{V}}_{\mathrm{Ca}{\left(\mathrm{OH}\right)}_2 0,5\mathrm{M}}=2 \mathrm{lít}\end{array}$