Đáp án đúng: D
Giải chi tiết:\(4este + 0,17mol\,{H_2} \to Y\left| \begin{gathered} \xrightarrow{{ + 0,11mol\,NaOH}}\left\{ \begin{gathered} {C_n}{H_{2n - 4}}{O_4}N{a_2}:0,03mol \hfill \\ {C_m}{H_{2m - 1}}{O_2}Na:0,05mol \hfill \\ \end{gathered} \right. + 6,88gT\left\{ {{C_a}{H_{2a + 1}}OH} \right. \hfill \\ \xrightarrow{{ + 0,805mol\,{O_2}}}\left\{ \begin{gathered} C{O_2}:a \hfill \\ {H_2}O:b \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right.\)
0,01 mol X cần 0,09 mol O2 => 0,08 mol X cần 0,72 mol O2
=> Đốt Y cần 0,72 + 0,17:2 = 0,805 mol O2
nCOO(Y) = 0,11 mol => nO(Y) = 0,22 mol. Đốt Y thu được nCO2 = a mol; nH2O = b mol
BT “O”: 2a + b = 0,22 + 0,805.2 = 1,83
Trong Y gồm 
\(\left\{ \begin{gathered} Este\,hai\,chuc(x\,mol) \hfill \\ Este\,don\,chuc(y\,mol) \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} 4x + 2y = 0,22 \hfill \\ x + y = 0,08 \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} x = 0,03 \hfill \\ y = 0,05 \hfill \\ \end{gathered} \right.\)
=> nCO2 – nH2O= 0,03 => 
\(\left\{ \begin{gathered} 2a + b = 1,83 \hfill \\ a - b = 0,03 \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} a = 0,62 \hfill \\ b = 0,59 \hfill \\ \end{gathered} \right.\)
=> mY = 0,62.44 + 0,89.18 – 0,805.32 = 12,14 (g)
=> m muối = 12,14 + 0,11.40 – 6,88 = 9,66 (g)
\(\begin{gathered} {n_{ancol}}{\text{ }} = {\text{ }}{n_{NaOH{\text{ }}}} = {\text{ }}0,11{\text{ }} \to {\text{ }}\overline M {\text{ }} = {\text{ }}\frac{{6,88}}{{0,11}}{\text{ }} = {\text{ }}14n{\text{ }} + {\text{ }}18{\text{ }} \to {\text{ }}\overline n = 3,182 \to {n_C} = 0,35(mol) \hfill \\ \to {n_{C(muoi)}} = 0,62 - 0,35 = 0,27(mol) \to \overline C = 3,375 \hfill \\ \end{gathered} \)
Ta có:
3                      0,625               3
            3,375               =
4                      0,375               5
=> muối là C4H4O4Na2 (0,03 mol)
=> m = 4,86 gam => %m muối = 50,31%
Đáp án D