Đề kiểm tra cuối học kì 2 môn Hoá 9 có đáp án (Mới nhất) - Đề 8

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{D} \\ {\mathrm{n}}_{\mathrm{rượu}} =\frac{9,2}{46} = 0,2 \left(\mathrm{mol}\right) \\ {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH} + 3{\mathrm{O}}_2 \stackrel{{\mathrm{t}}^{\mathrm{o}}}{\to } 2{\mathrm{CO}}_2 + 3{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Theo} \mathrm{phương} \mathrm{trình} \mathrm{phản} \mathrm{ứng}: {\mathrm{n}}_{{\mathrm{O}}_2} = 3 {\mathrm{n}}_{\mathrm{rượu}} = 3.0,2 = 0,6 \left(\mathrm{mol}\right) \\ {\mathrm{V}}_{{\mathrm{O}}_2}= 0,6.22,4 = 13,44 \left(\mathrm{l}\right) \\ \mathrm{Do} \mathrm{trong} \mathrm{không} \mathrm{khí}, \mathrm{oxi} \mathrm{chiếm} \mathrm{khoảng} 20\% \mathrm{về} \mathrm{thể} \mathrm{tích}. \\ \to \mathrm{Vkk} = \frac{{\mathrm{V}}_{{\mathrm{O}}_2}}{20\mathrm{\%}} = \frac{13,44}{20\mathrm{\%}} = 67,2 \left(\mathrm{l}\right) \end{gathered}$$

Đáp án đúng là: B

C₆H₁₂O₆ $\stackrel{men}{\to }$2CO₂ + C₂H₅OH

Vậy A là CO₂

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{B} \\ {\mathrm{n}}_{{\mathrm{O}}_2} = \frac{\mathrm{8,96}}{\mathrm{22,4}}= 0,4 \left(\mathrm{mol}\right) \\ 2{\mathrm{C}}_2{\mathrm{H}}_2 + 5{\mathrm{O}}_2 \stackrel{{\mathrm{t}}^{\mathrm{o}}}{\to } 4{\mathrm{CO}}_2 + 2{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Theo} \mathrm{phương} \mathrm{trình} \mathrm{phản} \mathrm{ứng}: {\mathrm{n}}_{{\mathrm{C}}_2{\mathrm{H}}_2} = \frac{2}{5}{\mathrm{n}}_{{\mathrm{O}}_2} = \frac{2}{5}.0,4 = 0,16 \left(\mathrm{mol}\right) \\ {\mathrm{m}}_{{\mathrm{C}}_2{\mathrm{H}}_2} = 0,16.26 = 4,16 \left(\mathrm{g}\right) \end{gathered}$$

$$\begin{gathered} \mathbf{a}\mathbf{)} {\mathrm{CH}}_2 = {\mathrm{CH}}_2 + {\mathrm{Br}}_2 \to \mathrm{Br} – {\mathrm{CH}}_2 – {\mathrm{CH}}_2 – \mathrm{Br} (1) \\ \mathbf{b}\mathbf{)} {\mathrm{n}}_{{\mathrm{Br}}_2} = \frac{80}{160} = 0,5 \left(\mathrm{mol}\right) \\ \mathrm{Theo} \mathrm{phương} \mathrm{trình} \mathrm{phản} \mathrm{ứng} \left(1\right): {\mathrm{n}}_{{\mathrm{C}}_{{}_2}{\mathrm{H}}_4} = {\mathrm{n}}_{{\mathrm{Br}}_2} = 0,5 \left(\mathrm{mol}\right) \\ {\mathrm{m}}_{{\mathrm{C}}_{{}_2}{\mathrm{H}}_4}= 0,5.28 = 14 \left(\mathrm{g}\right) \\ \% {\mathrm{m}}_{{\mathrm{C}}_{{}_2}{\mathrm{H}}_4}=\frac{14}{28} .100\% = 50\% \\ \% {\mathrm{m}}_{{\mathrm{CH}}_4}= 100\% - \%{\mathrm{m}}_{{\mathrm{C}}_{{}_2}{\mathrm{H}}_4} = 100\% - 50\% = 50\% \end{gathered}$$

$$\begin{gathered} {\mathrm{n}}_{{\mathrm{CO}}_2} =\frac{0,672}{22,4} = 0,03 \left(\mathrm{mol}\right) \to {\mathrm{n}}_{\mathrm{C}} = {\mathrm{n}}_{{\mathrm{CO}}_2} = 0,03 \left(\mathrm{mol}\right) \\ {\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}} = \frac{0,27}{18} = 0,015 \left(\mathrm{mol}\right) \to {\mathrm{n}}_{\mathrm{H}} = 2{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}} = 0,015.2 = 0,03 \left(\mathrm{mol}\right) \\ {\mathrm{m}}_{\mathrm{O}\left(\mathrm{A}\right)} = {\mathrm{m}}_{\mathrm{A}} – {\mathrm{m}}_{\mathrm{C}} – {\mathrm{m}}_{\mathrm{H}} = 0,39 – 0,03.12 – 0,03.1 = 0 \left(\mathrm{g}\right) \\ \to \mathrm{A} \mathrm{không} \mathrm{chứa} \mathrm{oxi} \\ \mathrm{Gọi} \mathrm{công} \mathrm{thức} \mathrm{phân} \mathrm{tử} \mathrm{của} \mathrm{A} \mathrm{có} \mathrm{dạng} {\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{y}} \\ \mathrm{x}: \mathrm{y} = {\mathrm{n}}_{\mathrm{C}}: {\mathrm{n}}_{\mathrm{H}} = 0,03 : 0,03 = 1 : 1 \\ \to \mathrm{x} = \mathrm{y}\to \mathrm{Công} \mathrm{thức} \mathrm{phân} \mathrm{tử} \mathrm{của} \mathrm{A} \mathrm{có} \mathrm{dạng} {\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{x}} \\ \mathrm{Ta} \mathrm{có}: 25 < {\mathrm{M}}_{\mathrm{A}} = 13\mathrm{x} < 30 \\ \to 1,92 < \mathrm{x} < 2,31 \\ \to \mathrm{x} = 2 \\ \to \mathrm{Công} \mathrm{thức} \mathrm{phân} \mathrm{tử} \mathrm{của} \mathrm{A} \mathrm{là} {\mathrm{C}}_2{\mathrm{H}}_2 \end{gathered}$$