Đốt cháy hoàn toàn 4,4 gam một hợp chất hữu cơ X (CnH2nO2), thu được V lít khí

* Hướng dẫn giải

$$\begin{gathered} \mathbf{a}\mathbf{)}\mathbf{ }{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}} = \frac{3,6}{18} = 0,2 (\mathrm{mol}) \\ {\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}}{\mathrm{O}}_2 + \frac{3\mathrm{n}-2}{2}{\mathrm{O}}_2 \stackrel{{\mathrm{t}}^{\mathrm{o}}}{\to } {\mathrm{nCO}}_2 + {\mathrm{nH}}_2\mathrm{O} \\ \mathrm{Theo} \mathrm{phương} \mathrm{trình} \mathrm{phản} \mathrm{ứng}: {\mathrm{n}}_{\mathrm{X}} =\frac{1}{\mathrm{n}}{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}= \frac{1}{\mathrm{n}} .0,2 (\mathrm{mol}) \\ {\mathrm{M}}_{\mathrm{X}} = \frac{4,4.\mathrm{n}}{0,2} = 22\mathrm{n} = 12.\mathrm{n} + 2\mathrm{n}.1 + 16.2 \\ \to \mathrm{n} = 4 \\ \to \mathrm{Công} \mathrm{thức} \mathrm{phân} \mathrm{tử} \mathrm{của} \mathrm{X} \mathrm{là} {\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_2 \end{gathered}$$

$$\begin{gathered} \mathbf{b}\mathbf{)} {\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_2 + 5{\mathrm{O}}_2 \stackrel{{\mathrm{t}}^{\mathrm{o}}}{\to } 4{\mathrm{CO}}_2 + 4{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Theo} \mathrm{phương} \mathrm{trình} \mathrm{phản} \mathrm{ứng}: {\mathrm{n}}_{{\mathrm{CO}}_2}= {\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}} = 0,2 (\mathrm{mol}) \\ {\mathrm{n}}_{\mathrm{Ba}{\left(\mathrm{OH}\right)}_2}\mathbf{ }\mathbf{ }= 0,15.1 = 0,15 (\mathrm{mol}) \\ \mathrm{Đặt} \mathrm{T} = \frac{{\mathrm{n}}_{{\mathrm{CO}}_2}}{{\mathrm{n}}_{\mathrm{Ba}{\left(\mathrm{OH}\right)}_2}} = \frac{0,2}{0,15} = 1,33 \\ \to 1 < \mathrm{T} < 2 \\ \to \mathrm{Tạo} \mathrm{hai} \mathrm{muối} {\mathrm{BaCO}}_3 \mathrm{và} \mathrm{Ba}{\left({\mathrm{HCO}}_3\right)}_2 \mathrm{Ba}{\left(\mathrm{OH}\right)}_2 + {\mathrm{CO}}_2 \to {\mathrm{BaCO}}_3 + {\mathrm{H}}_2\mathrm{O} (1) \\ \mathrm{Ba}{\left(\mathrm{OH}\right)}_2 + 2{\mathrm{CO}}_2 \to \mathrm{Ba}{\left({\mathrm{HCO}}_3\right)}_2 (2) \\ \mathrm{Đặt} \mathrm{x} \mathrm{và} \mathrm{y} \mathrm{lần} \mathrm{lượt} \mathrm{là} \mathrm{số} \mathrm{mol} \mathrm{Ba}{\left(\mathrm{OH}\right)}_2 \mathrm{ở} \mathrm{phương} \mathrm{trình} (1) \mathrm{và} (2) \\ {\mathrm{n}}_{\mathrm{Ba}{\left(\mathrm{OH}\right)}_2} = \mathrm{x} + \mathrm{y} = 0,15 (\mathrm{a}) \\ {\mathrm{n}}_{{\mathrm{CO}}_2}= \mathrm{x} + 2\mathrm{y} = 0,2 (\mathrm{b}) \\ \mathrm{Từ} (1) \mathrm{và} (2) \mathrm{suy} \mathrm{ra}: \mathrm{x} = 0,1 \mathrm{và} \mathrm{y} = 0,05 \\ \mathrm{Theo} \mathrm{phương} \mathrm{trình} (1): {\mathrm{n}}_{{\mathrm{BaCO}}_3}= {\mathrm{n}}_{\mathrm{Ba}{\left(\mathrm{OH}\right)}_2} (1) = 0,1 (\mathrm{mol}) \\ {\mathrm{m}}_{{\mathrm{BaCO}}_3}= 0,1.197 = 19,7 (\mathrm{g}) \end{gathered}$$

$$\begin{gathered} \mathbf{ }\mathbf{c}\mathbf{)}\mathbf{ }{\mathrm{n}}_{\mathrm{X}} = \frac{4,4}{88} = 0,05 (\mathrm{mol}) \\ \mathrm{Hợp} \mathrm{chất} \mathrm{hữu} \mathrm{cơ} \mathrm{X} \mathrm{chứa} \mathrm{C}, \mathrm{H}, \mathrm{O} \mathrm{tác} \mathrm{dụng} \mathrm{với} \mathrm{NaOH} \mathrm{thu} \mathrm{được} \mathrm{muối} \\ \to \mathrm{X} \mathrm{là} \mathrm{axit} \mathrm{hữu} \mathrm{cơ} \mathrm{hoặc} \mathrm{este} \mathrm{RCOOR}\text{'} + \mathrm{NaOH} \to \mathrm{RCOONa} + \mathrm{R}\text{'}\mathrm{OH} \\ \mathrm{Theo} \mathrm{phương} \mathrm{trình} \mathrm{phản} \mathrm{ứng}: {\mathrm{n}}_{\mathrm{muối}} = {\mathrm{n}}_{\mathrm{X}} = 0,05 (\mathrm{mol}) \\ {\mathrm{M}}_{\mathrm{muối}} = \frac{4,1}{0,05} = 82 = \mathrm{R} + 12 + 16.2 + 23 \\ \to \mathrm{R} = 15 (- {\mathrm{CH}}_3) \\ \to \mathrm{R}\text{'} = 88 – 15 – 12 – 16.2 = 29 (- {\mathrm{C}}_2{\mathrm{H}}_5) \\ \mathrm{Công} \mathrm{thức} \mathrm{cấu} \mathrm{tạo} \mathrm{của} \mathrm{X} \mathrm{là} {\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5 \end{gathered}$$