Chọn đáp án C

Giải thích:

\(\begin{array}{l} + \,\,Ancol\,\,no,\,\,hai\,\,chuc,\,\,3C\,:\,{C_3}{H_6}{(OH)_2}.\\ + \,\,\left\{ \begin{array}{l} Axit\,\,no,\,\,don\,\,chuc:\,\,\underbrace {{C_n}{H_{2n}}{O_2}}_{1\pi } \to \underbrace {COO}_{1\pi } + \underbrace {{C_{n'}}{H_{2n' + 2}}}_{0\pi } \to COO + C{H_2} + {H_2} \Rightarrow {n_{{H_2}}} = {n_{axit}}\\ Ancol\,\,no:\,\,{C_3}{H_6}{(OH)_2} \Leftrightarrow {C_3}{H_8}{O_2} \to {C_3}{H_6} + {H_2}{O_2} \to C{H_2} + {H_2}{O_2} \Rightarrow {n_{{H_2}{O_2}}} = {n_{ancol}}\\ Este\,\,no,\,\,hai\,\,chuc:\underbrace {{C_m}{H_{2m - 4}}{O_4}}_{2\pi }\,\, \to \underbrace {2COO}_{2\pi } \to \underbrace {{C_{m'}}{H_{2m' + 2}}}_{0\pi } \to COO + C{H_2} + {H_2} \Rightarrow {n_{{H_2}}} = {n_{este}} \end{array} \right.\\ \Rightarrow X \to \left\{ \begin{array}{l} COO:\,\,0,1\,\,mol\,\,( = {n_{KOH}})\\ C{H_2}:\,\,x\,\,mol\\ ({H_2},\,{H_2}{O_2}):\,\,0,09\,\,mol\,\,( = {n_X}) \end{array} \right\} \to \left\{ \begin{array}{l} C{O_2}:\,\,(0,1 + x)\,\,mol\\ {H_2}O:\,\,(0,09 + x)\,\,mol \end{array} \right\}\\ \Rightarrow 44(0,1 + x) - 18(0,09 + x)\, = 10,84 \Rightarrow x = 0,31 \Rightarrow BTO:\,\,{n_{{H_2}{O_2}}} = 0,03 \Rightarrow {n_{{H_2}}} = 0,06 \Rightarrow {m_X} = 9,88.\\ \Rightarrow \left\{ \begin{array}{l} {n_{axit}} + 2{n_{este}} = {n_{COO}} = 0,1\\ {n_{axit}} + {n_{este}} = {n_{{H_2}}} = 0,06 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} {n_{este}} = 0,04\\ {n_{axit}} = 0,02 \end{array} \right. \Rightarrow X \to \left\{ \begin{array}{l} {n_{{H_2}O}} = {n_{axit}} = 0,02\\ {n_{{C_3}{H_6}{{(OH)}_2}}} = {n_{{C_3}{H_6}{{(OH)}_2}/X}} + {n_{este}} = 0,07 \end{array} \right.\\ \Rightarrow {m_{muoi}} = \underbrace {9,88}_{{m_X}} + \underbrace {0,1.56}_{{m_{KOH}}} - \underbrace {0,02.18}_{{m_{{H_2}O}}} - \underbrace {0,07.76}_{{m_{{C_3}{H_6}{{(OH)}_3}}}} = 9,8\,gam. \end{array}\)