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

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{D}\mathbf{ } \\ {\mathrm{n}}_{{\mathrm{rượu} \mathrm{etylic}}_{}}=\frac{4,6}{46} = 0,1 \left(\mathrm{mol}\right) \\ {\mathrm{n}}_{\mathrm{este}} = \frac{4,4}{88} = 0,05 \left(\mathrm{mol}\right) \\ {\mathrm{CH}}_3\mathrm{COOH} + {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH} \stackrel{{\mathrm{H}}_2{\mathrm{SO}}_4,{\mathrm{t}}^{\mathrm{o}}}{\rightleftarrows } {\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_{5 }+ {\mathrm{H}}_2\mathrm{O} \\ \mathrm{Theo} \mathrm{phương} \mathrm{trình} \mathrm{phản} \mathrm{ứng}: {\mathrm{n}}_{\mathrm{rượu} \mathrm{etylic} \mathrm{phản} \mathrm{ứng}} = {\mathrm{n}}_{\mathrm{este}} = 0,05 \left(\mathrm{mol}\right) \\ \mathrm{H} = \frac{{\mathrm{n}}_{\mathrm{pu}}}{{\mathrm{n}}_{\mathrm{bd}}} .100\% = \frac{0,05}{0,1} .100\% = 50\% \end{gathered}$$

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{C} \\ {\mathrm{n}}_{{\mathrm{CaCO}}_3} = \frac{10}{100} = 0,1 \left(\mathrm{mol}\right) \\ {\mathrm{CaCO}}_3 + 2\mathrm{HCl} \to {\mathrm{CaCl}}_2 + {\mathrm{H}}_2\mathrm{O} + {\mathrm{CO}}_2 \\ \mathrm{Theo} \mathrm{phương} \mathrm{trình} \mathrm{phản} \mathrm{ứng}: {\mathrm{n}}_{{\mathrm{CO}}_2}= {\mathrm{n}}_{{\mathrm{CaCO}}_3}= 0,1 \left(\mathrm{mol}\right) \\ {\mathrm{n}}_{{\mathrm{CO}}_2}= 0,1.22,4 = 2,24 \left(\mathrm{l}\right) \end{gathered}$$

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{B}\mathbf{ } \\ \%\mathrm{C} = \frac{12.2}{12.2+4.1+16.2} .100\% = 40\% \end{gathered}$$

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{D}\mathbf{ } \\ {\mathrm{CH}}_3\mathrm{COOH} + {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH} \stackrel{{\mathrm{H}}_2{\mathrm{SO}}_4,{\mathrm{t}}^{\mathrm{o}}}{\rightleftarrows } {\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5 + {\mathrm{H}}_2\mathrm{O} \end{gathered}$$

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{B}\mathbf{ } \\ {\left({\mathrm{C}}_6{\mathrm{H}}_{10}{\mathrm{O}}_5\right)}_{\mathrm{n}} + {\mathrm{nH}}_2\mathrm{O} \stackrel{{\mathrm{t}}^{\mathrm{o}},\mathrm{axit}}{\to } {\mathrm{nC}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6 \left(1\right) \\ {\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_{6 }+ {\mathrm{Ag}}_2\mathrm{O} \stackrel{{\mathrm{t}}^{\mathrm{o}},{\mathrm{NH}}_2}{\to } {\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_7 + 2\mathrm{Ag} \left(2\right) \\ {\mathrm{n}}_{\mathrm{Ag}} = \frac{21,6}{108} = 0,2 \left(\mathrm{mol}\right) \\ \mathrm{Theo} \mathrm{phương} \mathrm{trình} \left(2\right): {\mathrm{n}}_{\mathrm{glucozơ}} = \frac{1}{2} {\mathrm{n}}_{\mathrm{Ag}} =\frac{1}{2} .0,2 = 0,1 \left(\mathrm{mol}\right) \\ \mathrm{Theo} \mathrm{phương} \mathrm{trình} \left(1\right): {\mathrm{n}}_{\mathrm{tinh} \mathrm{bột}} = \frac{1}{\mathrm{n}} {\mathrm{n}}_{\mathrm{glucozơ}} = \frac{1}{\mathrm{n}} .0,1 \left(\mathrm{mol}\right) \\ {\mathrm{m}}_{\mathrm{tinh} \mathrm{bột}} = \frac{1}{\mathrm{n}}\mathrm{t} .0,1.(12.6 + 10.1 + 16.5).\mathrm{n} = 16,2 \left(\mathrm{g}\right) \end{gathered}$$

a) K₂CO₃ + Ca(OH)₂ $\to$CaCO₃ + 2KOH

b) C₂H₅OH + Na $\to$2C₂H₅ONa + H₂

c) CH₃COOH + KOH $\to$ CH₃COOK + H₂O

Dẫn ba khí metan, etilen và cacbonic qua dung dịch brom. Khí nào làm mất màu dung dịch brom là etilen.
CH₂ = CH₂ + Br – Br→Br – CH₂ – CH₂ – BrHai khí còn lại không làm mất màu dung dịch brom là metan và cacbonic.
Dẫn hai khí còn lại qua dung dịch nước vôi trong dư. Khí nào làm vẩn đục nước vôi trong là cacbonic.
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

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

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

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