Hỗn hợp X gồm 3 este đơn chức, tạo thành từ cùng một ancol Y với 3 axit cacboxylic (phân tử chỉ có nhóm -COOH); trong đó, có hai axit no

Trả lời

$n_{\text{COO}}=n_{\text{OH}}=2n_{{\text{H}}_2}=0,096 \text{mol}$

$m_{\text{ancal }}=2,976+m_{H_2}=2,976+0,048.2=3,072\text{gam}$

 $\to M_{\text{ancal }}=\frac{3,072}{0,096}=32\to {\text{CH}}_3\text{OH}$

Vì đốt cháy  $5,88\text{ gam X nên }{n}_{COO}=\frac{0,096.5,88}{7,056}=0,08 \text{mol}$

$n_{{\text{H}}_2\text{O}}=0,22 \text{mol}\to n_{{\text{CO}}_3}=n_C=\frac{5,88-0,08.32-0,22.2}{12}=0,24 \text{mol}$

$\overline{C}=\frac{0,24}{0,08}=3\to \left\{\begin{array}{l}{\text{HCOOCH}}_3(k=1)\\ {\text{CH}}_3{\text{COOCH}}_3(k=1)\\ {\text{C}}_n{\text{H}}_{2n-1}{\text{COOCH}}_3(k=2,n⩾3)\end{array}\right.$

Ta có:

$(k-1)n_{\text{hchc }}=n_{{\text{CO}}_3}-n_{{\text{H}}_2\text{O}}\to (1-1)n_{\text{este no }}+(2-1).n_{\text{este khongno }}=0,24-0,22\to n_{estekhongno}=0,02$

$\to n_{\text{este }no}=0,06 \text{mol}$

Quy đổi $\left\{\begin{array}{l}{\text{HCOOCH}}_3(0,06 \text{mol})\\ {\text{C}}_3{\text{H}}_5{\text{COOCH}}_3(0,02 \text{mol})\\ {\text{CH}}_2(0,24-0,06.2-0,02.5=0,02 \text{mol})\end{array}\right.$

$\to \left\{\begin{array}{l}\text{HCOOCH}(0,04 \text{mol})\\ {\text{CH}}_3{\text{COOCH}}_3(0,02 \text{mol})\to {\text{mC}}_3{\text{H}}_5{\text{COOCH}}_3=2\text{gam}\\ {\text{C}}_3{\text{H}}_5{\text{COOCH}}_3(0,02 \text{mol})\end{array}\right.$