X, Y là hai este đều đơn chức, cùng dãy

Chọn đáp án C

* Ta xác định công thức 2 ancol trước:

Hệ $\left\{\begin{array}{l}a+b=\mathrm{0,24}\\ 32a+46b=\mathrm{8,8}\end{array}\right.\Rightarrow \left\{\begin{array}{l}a=\mathrm{0,16}\\ b=\mathrm{0,08}\end{array}\right.$

Sơ đồ: $–COO{–}_{este} + NaOH \to –COON{a}_{muoi} + –O{H}_{ancol}$

$mol\mathrm{0,24}\mathrm{0,}24\mathrm{0,}24\stackrel{}{\leftarrow }\mathrm{0,24}$ 

3 este có dạng $\left\{\begin{array}{l}C{H}_{3}COOC{H}_3\\ C{H}_{3}COO{C}_2{H}_5\\ C{H}_2\left(COO{R}^{\text{'}}\right)\left(COO{R^{\text{'}}}^{\text{'}}\right)\end{array}\right.$

($R^{\text{'}}$ và $R^{\text{'}\text{'}}$ là CH₃ hoặc C₂H₅, có thể giống hoặc khác nhau)

2 muối là $\left\{\begin{array}{l}C{H}_{3}COONa:cmol\\ C{H}_2{\left(COONa\right)}_2:dmol\end{array}\right.\Rightarrow n_{-COONa}=c+2d=\mathrm{0,24}$

$C{H}_{3}COONa+NaOH\underset{t°}{\overset{CaO}{\to }}C{H}_4+N{a}_{2}C{O}_3$

mol c        →     c                  c        c

$C{H}_2{\left(COONa\right)}_2+2NaOH\underset{t°}{\overset{CaO}{\to }}C{H}_4+2N{a}_{2}C{O}_3$

mol d     →            2d                 d             2d

$\Rightarrow m_{2muoi}=m_{C{H}_4}+m_{N{a}_{2}C{O}_3}-m_{NaOH}=m+106.\left(c+2d\right)-40.\left(c+2d\right)=$

$=m+\mathrm{106.0,24}-\mathrm{40.0,24}=m+\mathrm{15,84}$

– BTKL cho phản ứng thủy phân:

$m_E+m_{NaOH}=n_{muoái}+n_{ancol}\iff \mathrm{5,7}m+\mathrm{40.0,24}=m+\mathrm{15,84}+\mathrm{8,8}\Rightarrow m=\mathrm{3,2}gam$

Hệ $\left\{\begin{array}{l}c+2d=\mathrm{0,24}\\ n_{C{H}_4}=c+d=\frac{\mathrm{3,2}}{16}=\mathrm{0,2}\end{array}\right.\Rightarrow \left\{\begin{array}{l}c=\mathrm{0,16}\\ d=\mathrm{0,04}\end{array}\right.$  

E gồm $\left\{\begin{array}{l}\left.\begin{array}{l}C{H}_{3}COOC{H}_3:\text{x }mol\\ C{H}_{3}COO{C}_2{H}_5:ymol\end{array}\right\}\Rightarrow x+y=\mathrm{0,16}\\ C{H}_2\left(COO{R}^{\text{'}}\right)\left(COO{R}^{\text{'}\text{'}}\right)=\mathrm{0,04}mol\end{array}\right.$

– Nếu$R^{\text{'}}$ và $R^{\text{'}\text{'}}$cùng là $C{H}_3\Rightarrow \left\{\begin{array}{l}x+y=\mathrm{0,16}\\ 74x+88y+\mathrm{132.0,04}=\mathrm{5,7.3,2}=\mathrm{18,24}\end{array}\right.$

$\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,08}\\ y=\mathrm{0,08}\end{array}\right.$ (thỏa mãn n_Y > n_Z) $\Rightarrow m_Z=\mathrm{0,04.132}=\mathrm{5,28}gam\Rightarrow$ chọn C

– Nếu R' là CH₃. R'' là $C_2{H}_5\Rightarrow \left\{\begin{array}{l}x+y=\mathrm{0,16}\\ 74x+88y+\mathrm{146.0,04}=\mathrm{5,7.3,2}=\mathrm{18,24}\end{array}\right.$

$\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,12}\\ y=\mathrm{0,04}\end{array}\right.$ (loai, vì không thỏa mãn n_Y > n_Z)

– Nếu R' và R'' cùng là $C_2{H}_5\Rightarrow \left\{\begin{array}{l}x+y=\mathrm{0,16}\\ 74x+88y+\mathrm{160.0,04}=\mathrm{5,7.3,2}=\mathrm{18,24}\end{array}\right.$

$\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,16}\\ y=0\end{array}\right.$(loại)