X, Y (MX < MY) là hai chất hữu cơ kế tiếp thuộc dãy đồng đẳng ancol anlylic; Z là axit no, hai

Trả lời

$n_{{\text{O}}_2}=0,38 \text{mol};n_{{\text{H}}_2\text{O}}=0,255 \text{mol}$ 

Bảo toàn khối lượng: $m_{{\text{CO}}_2}=6,95+0,38.32-4,59=14,52\text{gam}\to n_{{\text{CO}}_2}=0,33 \text{mol}$

Bảo toàn nguyên tố oxi:  $n_{O\text{ trong }E}=\frac{0,255+0,33.2-0,38.2}{2}=0,155 \text{mol}$

$\left\{\begin{array}{l}{k}_{X,Y}=1k_Z=2\left(2COOH\right)\\ k_Z=2\left(2COOH\right)\\ k_T=4(2COOH+2lkC=C)\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_Z+3n_T=n_{C{O}_2}-n_{H_{2}O}=0,075\\ n_X+2_T=n_{B{r}_2}=0,055\\ n_{X_{1}Y}+n_Z+n_T=0,155\end{array}\right.$

$\to \left\{\begin{array}{l}{n}_{X,Y}=0,015\\ n_Z=0,015\\ n_T=0,02\end{array}\right.$

$\left\{\begin{array}{l}{C}_X⩾3;C_Y⩾4\\ C_Z⩾2\\ C_T⩾10\end{array}\right.\Rightarrow \left\{\begin{array}{l}\Delta C{H}_2=0,33-0,015.3-0,015.3-0,02.10=0,04>n_Z+n_T\\ 0,04\text{ mol }C{H}_2=0,02+0,015+0,005\end{array}\right.$

E gồm

$\left\{\begin{array}{l}{\text{C}}_3{\text{H}}_5\text{OH}(0,01 \text{mol})\\ {\text{C}}_4{\text{H}}_7\text{OH}(0,005 \text{mol})\\ {\text{CH}}_2{\text{CH}}_2(0,015 \text{mol})\\ {\text{C}}_3{\text{H}}_5{\text{OOCCC}}_2{\text{H}}_4{\text{COOC}}_4{\text{H}}_7(0,02 \text{mol})\end{array}\right.$

${\text{C}}_3{\text{H}}_5{\text{OOCC}}_2{\text{H}}_4{\text{COOC}}_4{\text{H}}_7(0,02 \text{mol})\stackrel{+{\text{O}}_2}{\to }0,22 {\text{molCO}}_2\stackrel{+\text{Ca}{\left(\text{OH}\right)}_2\text{du}}{\to }0,22 {\text{molCaCO}}_3$

$\to m_{{\text{CaCO}}_{\text{U}}}=22\text{gam}$