Nung m gam hỗn hợp X gồm KHCO3 và CaCO3 ở nhiệt độ cao đến khối lượng không đổi,

* Hướng dẫn giải

$$\begin{gathered} \underset{m\left(gam\right)}{\underbracket{hh\left\{\begin{array}{l}KHC{O}_3:\mathrm{0,8}m\\ CaC{O}_3:\mathrm{0,2}m\end{array}\right.}}\stackrel{t^o}{\to }ran\text{\hspace{0.17em}}Y\left\{\begin{array}{l}{K}_{2}C{O}_3\\ CaO\end{array}\right.\stackrel{H_{2}O}{\to }\left\{\begin{array}{l}ranZ:CaC{O}_3:\mathrm{0,2}m\\ ddE:\left\{\begin{array}{l}KOH\\ K_{2}C{O_3}_{d­}\end{array}\right.\stackrel{HCl1M}{\to }\left\{\begin{array}{l}Bat dau thoat khi:V_{1}lit\\ Khi thoat het:V_{2}lit\end{array}\right.\end{array}\right. \\ Vì dd E tác dụng với HCl tạo khí \Rightarrow K_{2}C{O}_3 dư \Rightarrow BT \left(Ca\right) \Rightarrow mCaC{O}_3 = 0,2m \left(gam\right) \Rightarrow m_{KHCO3} = 0,8m \left(gam\right) \end{gathered}$$

$Vì M_{CaCO3} = M_{KHCO3} \frac{n_{KHCO3}}{n_{CaCO3}}=\frac{\mathrm{0,8}m}{\mathrm{0,2}m}=4\Rightarrow Chon\left\{\begin{array}{l}{n}_{KHC{O}_3}=4mol\\ n_{CaC{O}_3}=1mol\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{K_{2}C{O}_3}=2mol\\ n_{CaO}=1mol\end{array}\right.$

$$\begin{gathered} CaO + H_{2}O \to Ca{\left(OH\right)}_2 \\ 1 mol \to 1 mol \\ Ca{\left(OH\right)}_2 + K_{2}C{O}_3 \to CaC{O}_3 + 2KOH \\ 1 mol < 2 mol \to 2 mol \\ Dư: 1 mol \end{gathered}$$

$$\begin{gathered} PTHH \left(1\right) O{H}^{-} + H^{+} \to H_{2}O \\ 2 2 \\ \left(2\right) C{O_3}^{2-} + H^{+} \to HC{O_3}^{-} \\ 1 1 1 \\ \left(3\right) HC{O_3}^{-} + H^{+} \to C{O}_2 + H_{2}O \\ 1 1 \\ \Rightarrow \frac{V_1}{V_2}=\frac{n_{H^{+}\left(\mathrm{1,2}\right)}}{n_{H^{+}\left(\mathrm{1,2,3}\right)}}=\frac{3}{4} \end{gathered}$$