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Câu hỏi: Cho các phản ứng xảy ra theo sơ đồ sau
$\begin{array}{*{35}{l}}
\left( 1 \right)~{{X}_{1}}+{{H}_{2}}O\xrightarrow[\text{ co }\!\!\grave{\mathrm{u}}\!\!\text{ mang nga }\!\!\hat{\mathrm{e}}\!\!\text{ n}]{\tilde{n}ie\ddot{a}n\text{ pha }\!\!\hat{\mathrm{a}}\!\!\text{ n}}{{X}_{2}}+Y\uparrow +Z\uparrow \\
\left( 2 \right)C{{O}_{2}}+{{X}_{2}}\to ~{{X}_{3}}. \\
\left( 3 \right)C{{O}_{2}}+2{{X}_{2}}\to {{X}_{4}}+{{H}_{2}}O. \\
\left( 4 \right){{X}_{3}}+{{X}_{5}}\to T+{{X}_{2}}+{{H}_{2}}O. \\
\left( 5 \right)2{{X}_{3}}+{{X}_{5}}\to T+{{X}_{4}}+2{{H}_{2}}O. \\
\end{array}$
Hai chất X2, X5 lần lượt là:
A. K2CO3, BaCl2.
B. KOH, Ba(HCO3)2.
C. KHCO3, Ba(OH)2.
D. KOH, Ba(OH)2.
$\begin{array}{*{35}{l}}
\left( 1 \right)~{{X}_{1}}+{{H}_{2}}O\xrightarrow[\text{ co }\!\!\grave{\mathrm{u}}\!\!\text{ mang nga }\!\!\hat{\mathrm{e}}\!\!\text{ n}]{\tilde{n}ie\ddot{a}n\text{ pha }\!\!\hat{\mathrm{a}}\!\!\text{ n}}{{X}_{2}}+Y\uparrow +Z\uparrow \\
\left( 2 \right)C{{O}_{2}}+{{X}_{2}}\to ~{{X}_{3}}. \\
\left( 3 \right)C{{O}_{2}}+2{{X}_{2}}\to {{X}_{4}}+{{H}_{2}}O. \\
\left( 4 \right){{X}_{3}}+{{X}_{5}}\to T+{{X}_{2}}+{{H}_{2}}O. \\
\left( 5 \right)2{{X}_{3}}+{{X}_{5}}\to T+{{X}_{4}}+2{{H}_{2}}O. \\
\end{array}$
Hai chất X2, X5 lần lượt là:
A. K2CO3, BaCl2.
B. KOH, Ba(HCO3)2.
C. KHCO3, Ba(OH)2.
D. KOH, Ba(OH)2.
$\begin{aligned}
& \begin{array}{*{35}{l}}
\left( 1 \right)2KC1+2{{H}_{2}}O\xrightarrow[co\text{ mang ngan}]{dien\text{ phan}}2KOH+C{{l}_{2}}+{{H}_{2}} \\
\left( 2 \right)C{{O}_{2}}+KOH\to ~KHC{{O}_{3}} \\
\left( 3 \right)C{{O}_{2}}+2KOH\to ~{{K}_{2}}C{{O}_{3}}+{{H}_{2}}O \\
\left( 4 \right)KHC{{O}_{3}}+Ba{{\left( OH \right)}_{2}}~\to BaC{{O}_{3}}+KOH+{{H}_{2}}O \\
\end{array} \\
& \left( 5 \right)2KHC{{O}_{3}}+Ba{{\left( OH \right)}_{2}}~\to BaC{{O}_{3}}+{{K}_{2}}C{{O}_{3}}+2{{H}_{2}}O \\
\end{aligned}$
=> X2 là KOH và X5 là Ba(OH)2
& \begin{array}{*{35}{l}}
\left( 1 \right)2KC1+2{{H}_{2}}O\xrightarrow[co\text{ mang ngan}]{dien\text{ phan}}2KOH+C{{l}_{2}}+{{H}_{2}} \\
\left( 2 \right)C{{O}_{2}}+KOH\to ~KHC{{O}_{3}} \\
\left( 3 \right)C{{O}_{2}}+2KOH\to ~{{K}_{2}}C{{O}_{3}}+{{H}_{2}}O \\
\left( 4 \right)KHC{{O}_{3}}+Ba{{\left( OH \right)}_{2}}~\to BaC{{O}_{3}}+KOH+{{H}_{2}}O \\
\end{array} \\
& \left( 5 \right)2KHC{{O}_{3}}+Ba{{\left( OH \right)}_{2}}~\to BaC{{O}_{3}}+{{K}_{2}}C{{O}_{3}}+2{{H}_{2}}O \\
\end{aligned}$
=> X2 là KOH và X5 là Ba(OH)2
Đáp án D.