Đốt cháy hoàn toàn 6,46 gam hỗn

Chọn đáp án C

Giải thích:

$\begin{array}{l}+\left\{\begin{array}{l}BTKL:\underset{\mathrm{6,46}}{\underbrace{m_E}}+32\underset{\mathrm{0,235}}{\underbrace{n_{O_2}}}=44\underset{\mathrm{0,24}}{\underbrace{n_{C{O}_2}}}+18\underset{?}{\underbrace{n_{H_{2}O}}}\\ BTO:n_{O/E}+2\underset{\mathrm{0,235}}{\underbrace{n_{O_2}}}=2\underset{\mathrm{0,24}}{\underbrace{n_{C{O}_2}}}+\underset{?}{\underbrace{n_{H_{2}O}}}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{H_{2}O}=\mathrm{0,19}\\ n_{O/E}=\mathrm{0,2}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{-COO-}=\mathrm{0,1}\\ n_{-COONa}=\mathrm{0,1}\\ n_{NaOHdu}=20\%\mathrm{.0,1}=\mathrm{0,02}\end{array}\right.\\ +T\stackrel{quydoi}{\to }\left\{\begin{array}{l}COONa:\mathrm{0,1}\\ NaOH:\mathrm{0,02}\\ C\\ H\end{array}\right\}\stackrel{O_2,t^o}{\to }\left\{\begin{array}{l}BTNa:N{a}_{2}C{O}_3:\mathrm{0,06}mol\\ H_{2}O:\mathrm{0,01}mol\\ C{O}_2:?\end{array}\right\}\Rightarrow n_{H/H_{2}O}=n_{H/NaOH}\end{array}$

$\begin{array}{l}\Rightarrow GocaxitkhongcoH\Rightarrow Muoila{\left(COONa\right)}_2:\mathrm{0,05}mol.\\ +Matkhac:X,Y,Zno,machho\Rightarrow 2ancollaphaidonchuc\Rightarrow \left\{\begin{array}{l}{n}_{2ancol}=n_{-COO-}=\mathrm{0,1}\\ {\overline{C}}_{ancol}=\frac{\mathrm{6,46}+\mathrm{0,1.40}-\mathrm{0,05.134}}{\mathrm{0,1}}=\mathrm{36,7}\end{array}\right.\\ \Rightarrow 2ancolla\left\{\begin{array}{l}C{H}_{3}OH\\ C_2{H}_{5}OH\end{array}\right\}\Rightarrow \left\{\begin{array}{l}Xla{\left(COOC{H}_3\right)}_2(M=118)\\ YlaC{H}_3{\left(COO\right)}_2{C}_2{H}_5(M=132\\ Zla{\left(COO{C}_2{H}_5\right)}_2(M=146)\end{array}\right\}\end{array}$