Đốt cháy hoàn toàn 0,1 mol một ancol đơn chức trong 0,7 mol

* Hướng dẫn giải

Đáp án D

Theo bảo toàn nguyên tố C, H, O, ta có :

$\left\{\begin{array}{l}{\mathrm{n}}_{{\mathrm{CO}}_2}=\mathrm{x}.{\mathrm{n}}_{{\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{y}}\mathrm{O}}; 2{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}={\mathrm{yn}}_{{\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{y}}\mathrm{O}}\\ {\mathrm{n}}_{{\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{y}}\mathrm{O}}+2{\mathrm{n}}_{{\mathrm{O}}_2 \mathrm{pư}}=2{\mathrm{n}}_{{\mathrm{CO}}_2}+{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}\\ {\mathrm{n}}_{{\mathrm{O}}_2 \mathrm{dư}}={\mathrm{n}}_{{\mathrm{O}}_2 \mathrm{bđ}}-{\mathrm{n}}_{{\mathrm{O}}_2 \mathrm{pư}}\end{array}\right.$

$$\begin{gathered} \left\{\begin{array}{l}{\mathrm{n}}_{{\mathrm{CO}}_2}=0,1\mathrm{x}; {\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}=0,05\mathrm{y}\\ {\mathrm{n}}_{{\mathrm{O}}_2 \mathrm{pư}}=0,1\mathrm{x}+0,025\mathrm{y}-0,05\\ {\mathrm{n}}_{{\mathrm{O}}_2 \mathrm{dư}}=0,75-0,1\mathrm{x}-0,025\mathrm{y}\end{array}\right. \\ \left\{\begin{array}{l}{\mathrm{n}}_{({\mathrm{CO}}_2,{\mathrm{H}}_2\mathrm{O},{\mathrm{O}}_2 \mathrm{dư})}=0,025\mathrm{y}+0,75=1\\ {\mathrm{n}}_{{\mathrm{O}}_2 \mathrm{pư}}=0,1\mathrm{x}+0,025\mathrm{y}-0,05<{\mathrm{n}}_{{\mathrm{O}}_2 \mathrm{bđ}}=0,7\end{array}\right. \\ \left\{\begin{array}{l}\mathrm{y}=10\\ \mathrm{x}<5\end{array}\Rightarrow \left\{\begin{array}{l}\mathrm{y}=10; \mathrm{x}=4\\ {\mathrm{m}}_{{\mathrm{C}}_4{\mathrm{H}}_{10}\mathrm{O}}=7,4 \mathrm{gam}\end{array}\right.\right. \end{gathered}$$