Cho 10,08 lít khí CO2 (đktc) vào 600 ml dung

* Hướng dẫn giải

Chọn đáp án C

$\underset{\mathrm{0,45}\text{\hspace{0.33em}mol}}{\underbrace{C{O}_2}}+\left\{\begin{array}{l}\underset{\mathrm{0,3}\text{ mol}}{\underbrace{KOH}}\\ \underset{\mathrm{0,48}\text{\hspace{0.33em}mol}}{\underbrace{{\text{K}}_{2}C{O}_3}}\end{array}\right.\stackrel{}{\to }\left|\begin{array}{l}\left\{\begin{array}{l}\underset{a\text{ mol}}{\underbrace{K_{2}C{O}_3}}\\ \underset{b\text{ mol}}{\underbrace{{\text{KHCO}}_3}}\end{array}\right.\stackrel{BaC{l}_2\text{\hspace{0.33em}d­}}{\to }\left|\begin{array}{l}\underset{a\text{\hspace{0.33em}mol}}{\underbrace{BaC{O}_3\downarrow }}\stackrel{t^0}{\to }\left|\begin{array}{l}\underset{a\text{ mol}}{\underbrace{BaO}}:m\text{ gam}\\ {\text{CO}}_2\end{array}\right.\\ \left\{\begin{array}{l}KCl\\ KHC{O}_3\end{array}\right.\end{array}\right.\\ H_{2}O\end{array}\right.$

$Thi\text{\hspace{0.33em}nghiem 1:}\left\{\begin{array}{l}\stackrel{BTNT.C}{\to }a+b=\mathrm{0,45}+\mathrm{0,48}\\ \stackrel{BTNT.K}{\to }2a+b=\mathrm{0,3}+\mathrm{2.0,48}\end{array}\right.\to \left\{\begin{array}{l}a=\mathrm{0,33}\\ b=\mathrm{0,6}\end{array}\right.$

$\to n_{BaO}\stackrel{BTNT.Ba}{=}{n}_{BaC{O}_3}\stackrel{BTNT.C\text{ cho thi nghiem 2}}{=}\mathrm{0,33}\text{ mol}\to \text{m = 153.0,33 = 50,49\hspace{0.33em}gam}$