Cho 3,36 lít khí CO2 (đktc) vào 400 ml dung dịch hỗn hợp KOH

* Hướng dẫn giải

$\underset{0,15 \mathrm{mol}}{\underbrace{C{O}_2}}+\left\{\begin{array}{l}\underset{0,1\mathrm{mol}}{\underbrace{KOH}}\\ \underset{0,16 \mathrm{mol}}{\underbrace{{\mathrm{K}}_{2}C{O}_3}}\end{array}\right.\to \left|\begin{array}{c}\left\{\begin{array}{l}\underset{a\mathrm{mol}}{\underbrace{K_{2}C{O}_3}}\\ \underset{b\mathrm{mol}}{\underbrace{{\mathrm{KHCO}}_3}}\end{array}\right.\\ H_{2}O\end{array}\stackrel{BaC{l}_2,d-}{\to }\right|\begin{array}{c}\underset{a \mathrm{mol}}{\underbrace{BaC{O}_3\downarrow }}\stackrel{t°}{\to }|\begin{array}{c}\underset{a\mathrm{mol}}{\underbrace{BaO}}:m\mathrm{gam}\\ {\mathrm{CO}}_2\end{array}\\ \left\{\begin{array}{l}KCl\\ KHC{O}_3\end{array}\right.\end{array}$

Thí nghiệm 1:

$$\begin{gathered} \left\{\begin{array}{l}\stackrel{BTNT.C}{\to }a+b=0,15+0,16\\ \stackrel{BTNT.K}{\to }2a+b=0,1+2.0,16\end{array}\right.\to \left\{\begin{array}{l}a=0,11\\ b=0,2\end{array}\right. \\ \to n_{BaO}\stackrel{BTNT.Ba}{=}{n}_{BaC{O}_3}\stackrel{BTNT.C cho thí nghiêm 2}{=}0,11\mathrm{mol}\to \mathrm{m}=153.0,11=16,83 \mathrm{gam} \end{gathered}$$

Chọn đáp án C.