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Câu hỏi: Đốt cháy 0,3 mol hỗn hợp X gồm metyl acrylat, etylen glicol, axetanđehit và ancol metylic cần dùng a mol O2. Sản phẩm cháy dẫn qua 400ml dung dịch Ba(OH)2 1M, lọc bỏ kết tủa, cho dung dịch Ca(OH)2 dư vào phần nước lọc thì thu được thêm 106,92 gam kết tủa nữa. Giá trị của a là bao nhiêu?
A. 0,43.
B. 1,25.
C. 0,91.
D. 0,75.
A. 0,43.
B. 1,25.
C. 0,91.
D. 0,75.
Quy đổi. $X\left\{ \begin{aligned}
& C{{H}_{2}}=CHCOOC{{H}_{3}}\Leftrightarrow {{C}_{4}}{{H}_{6}}{{O}_{2}}\Leftrightarrow {{C}_{4}}{{H}_{2}}.2{{H}_{2}}O \\
& C{{H}_{2}}OHC{{H}_{2}}OH\Leftrightarrow {{C}_{2}}{{H}_{6}}{{O}_{2}}\Leftrightarrow {{C}_{2}}{{H}_{2}}.2{{H}_{2}}O \\
& C{{H}_{3}}CHO\Leftrightarrow {{C}_{2}}{{H}_{4}}{{O}_{2}}\Leftrightarrow {{C}_{2}}{{H}_{2}}.{{H}_{2}}O \\
& C{{H}_{3}}OH\Leftrightarrow C{{H}_{4}}O\Leftrightarrow C{{H}_{2}}{{H}_{2}}O \\
\end{aligned} \right.\Rightarrow X:{{C}_{\overline{x}}}{{H}_{2}}.\overline{y}{{H}_{2}}O$
${{C}_{\overline{x}}}{{H}_{2}}.\overline{y}{{H}_{2}}O\xrightarrow{+{{O}_{2}}}\left\{ \begin{aligned}
& C{{O}_{2}} \\
& {{H}_{2}}O \\
\end{aligned} \right.\xrightarrow{+Ba{{\left( OH \right)}_{2}}}\left\{ \begin{aligned}
& BaC{{O}_{3}}\downarrow :x mol \\
& Ba{{\left( HC{{O}_{3}} \right)}_{2}}\xrightarrow{Ca{{\left( OH \right)}_{2}}}\left\{ \begin{aligned}
& BaC{{O}_{3}}:y mol \\
& CaC{{O}_{3}}:y mol \\
\end{aligned} \right. \\
\end{aligned} \right.$
$\left\{ \begin{aligned}
& {{n}_{Ba{{\left( OH \right)}_{2}}}}=x+y=0,4 \\
& {{m}_{\text{kết tủa}}}=100y+197y=106,92 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& y=0,36 \\
& x=0,04 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {{n}_{C{{O}_{2}}}}=0,76 \\
& \overline{x}=\dfrac{0,76}{0,3} \\
\end{aligned} \right.$
$\xrightarrow{BT\text{ electron}}\left( 4\overline{x}+2 \right){{n}_{X}}=4{{n}_{{{O}_{2}}}}\Rightarrow {{n}_{{{O}_{2}}}}=\left( \dfrac{0,76}{0,3}+\dfrac{1}{2} \right).0,3=0,91 mol$.
& C{{H}_{2}}=CHCOOC{{H}_{3}}\Leftrightarrow {{C}_{4}}{{H}_{6}}{{O}_{2}}\Leftrightarrow {{C}_{4}}{{H}_{2}}.2{{H}_{2}}O \\
& C{{H}_{2}}OHC{{H}_{2}}OH\Leftrightarrow {{C}_{2}}{{H}_{6}}{{O}_{2}}\Leftrightarrow {{C}_{2}}{{H}_{2}}.2{{H}_{2}}O \\
& C{{H}_{3}}CHO\Leftrightarrow {{C}_{2}}{{H}_{4}}{{O}_{2}}\Leftrightarrow {{C}_{2}}{{H}_{2}}.{{H}_{2}}O \\
& C{{H}_{3}}OH\Leftrightarrow C{{H}_{4}}O\Leftrightarrow C{{H}_{2}}{{H}_{2}}O \\
\end{aligned} \right.\Rightarrow X:{{C}_{\overline{x}}}{{H}_{2}}.\overline{y}{{H}_{2}}O$
${{C}_{\overline{x}}}{{H}_{2}}.\overline{y}{{H}_{2}}O\xrightarrow{+{{O}_{2}}}\left\{ \begin{aligned}
& C{{O}_{2}} \\
& {{H}_{2}}O \\
\end{aligned} \right.\xrightarrow{+Ba{{\left( OH \right)}_{2}}}\left\{ \begin{aligned}
& BaC{{O}_{3}}\downarrow :x mol \\
& Ba{{\left( HC{{O}_{3}} \right)}_{2}}\xrightarrow{Ca{{\left( OH \right)}_{2}}}\left\{ \begin{aligned}
& BaC{{O}_{3}}:y mol \\
& CaC{{O}_{3}}:y mol \\
\end{aligned} \right. \\
\end{aligned} \right.$
$\left\{ \begin{aligned}
& {{n}_{Ba{{\left( OH \right)}_{2}}}}=x+y=0,4 \\
& {{m}_{\text{kết tủa}}}=100y+197y=106,92 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& y=0,36 \\
& x=0,04 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {{n}_{C{{O}_{2}}}}=0,76 \\
& \overline{x}=\dfrac{0,76}{0,3} \\
\end{aligned} \right.$
$\xrightarrow{BT\text{ electron}}\left( 4\overline{x}+2 \right){{n}_{X}}=4{{n}_{{{O}_{2}}}}\Rightarrow {{n}_{{{O}_{2}}}}=\left( \dfrac{0,76}{0,3}+\dfrac{1}{2} \right).0,3=0,91 mol$.
Đáp án C.