Đề thi thử THPT môn Hóa năm 2022 có đáp án (Mới nhất) - Đề 10

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{A} \\ \begin{array}{l}{\mathrm{Na}}_2\mathrm{O}+{\mathrm{H}}_2\mathrm{O}\to 2\mathrm{NaOH}\\ 0,1 \to 0,2\end{array} \\ {\mathrm{C}}_{{\mathrm{M}}_{\mathrm{NaOH}}}=\frac{0,2}{0,2}=1\mathrm{M} \end{gathered}$$

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{D}\mathbf{ } \\ {\mathrm{HCOOCH}}_3 + \mathrm{NaOH} \stackrel{\mathrm{t}°}{\to }\mathrm{HCOONa} + {\mathrm{CH}}_3\mathrm{OH} \\ \to \mathrm{Sản} \mathrm{phẩm} \mathrm{thu} \mathrm{được} \mathrm{là} \mathrm{HCOONa} \mathrm{và} {\mathrm{CH}}_3\mathrm{OH} \end{gathered}$$

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{A}\mathbf{ } \\ {\left({\mathrm{C}}_6{\mathrm{H}}_{10}{\mathrm{O}}_5\right)}_{\mathrm{n}}\to {\mathrm{n}}_{\mathrm{C}} \end{gathered}$$₆H₁₂O₆→ 2nC2H5OH   162                                   →       46.2 3,2                                    →         m → mC2H5OHthực tế =75%.80%.3,24.46.2162=1,104 kg  → VC2H5OH   = 1,1040,8=1,38 lít →  Vdd C2H5OH =1,3820%=6,9 l lít

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{B}\mathbf{ }\mathbf{ }\mathbf{} \\ {\mathrm{n}}_{{\mathrm{CO}}_2}=0,18\mathrm{mol};{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}=0,21\mathrm{mol} \\ \to {\mathrm{m}}_{\mathrm{X}} = {\mathrm{m}}_{\mathrm{C}} + {\mathrm{m}}_{\mathrm{H}} = 2,58 \mathrm{gam} \\ \mathrm{Số} \mathrm{C} =\frac{{\mathrm{n}}_{{\mathrm{CO}}_2}}{{\mathrm{n}}_{\mathrm{X}}}=1,8 \\ \mathrm{Số} \mathrm{H} = \frac{2{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}}{{\mathrm{n}}_{\mathrm{X}}}=4,2 \\ \to \mathrm{k} = \frac{\left(2\mathrm{C}+2–\mathrm{H}\right)}{2}=0,7 \\ \to {\mathrm{n}}_{{\mathrm{Br}}_2}=\mathrm{k}.{\mathrm{n}}_{\mathrm{X}}=0,07 \mathrm{mol} \\ \mathrm{Tỉ} \mathrm{lệ}: 2,58 \mathrm{gam} \mathrm{X} \mathrm{phản} \mathrm{ứng} \mathrm{với} 0,07 \mathrm{mol} {\mathrm{Br}}_2 \\ \to 3,87 \mathrm{gam} \mathrm{X} \mathrm{phản} \mathrm{ứng} \mathrm{với} 0,105 \mathrm{mol} {\mathrm{Br}}_2 \end{gathered}$$

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{A}\mathbf{ } \\ {\mathrm{n}}_{{\mathrm{H}}_2} = 0,15\to {\mathrm{n}}_{\mathrm{Cl}-\left(\mathrm{muối}\right)} = 0,3 \mathrm{mol} \\ {\mathrm{m}}_{ \mathrm{muối} }= {\mathrm{m}}_{ \mathrm{kim} \mathrm{loại}} + {\mathrm{m}}_{\mathrm{Cl}-\left(\mathrm{muối}\right) } = 16,3 \mathrm{gam} \end{gathered}$$

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{B}\mathbf{ } \\ {\mathrm{n}}_{{\mathrm{C}}_2{\mathrm{H}}_3\mathrm{Cl}}=\frac{1000}{62,5}=16 \mathrm{mol} \\ \to {\mathrm{n}}_{{\mathrm{CH}}_4} \mathrm{phản} \mathrm{ứng} = 32 \mathrm{mol} \\ \to {\mathrm{V}}_{ \mathrm{khí} \mathrm{thiên} \mathrm{nhiên}} = \frac{32.22,4}{95\mathrm{\%}.15\mathrm{\%}.95\mathrm{\%}.90\mathrm{\%}}=5883{\mathrm{m}}^3 \end{gathered}$$

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{B}\mathbf{ } \\ \mathrm{Dung} \mathrm{dịch} \mathrm{sau} \mathrm{điện} \mathrm{phân} + \mathrm{Fe}\to \mathrm{NO} \mathrm{nên} \mathrm{chứa} {\mathrm{H}}^{+}, {{\mathrm{NO}}_3}^{-}\to \mathrm{Có} {\mathrm{O}}_2 \mathrm{ở} \mathrm{anot}. \mathrm{Chất} \mathrm{rắn} \mathrm{thu} \mathrm{được} \mathrm{gồm} 2 \mathrm{kim} \mathrm{loại} \mathrm{nên} \mathrm{có} {{\mathrm{Cu}}^2}^{+} \mathrm{dư} (\mathrm{x} \mathrm{mol}). \\ \mathrm{ne} =\frac{\mathrm{It}}{\mathrm{F}}=0,6 \\ \mathrm{Anot}: {\mathrm{n}}_{{\mathrm{Cl}}_2} = \mathrm{a} \mathrm{và} {\mathrm{n}}_{{\mathrm{O}}_2} = \mathrm{b} \\ \to \mathrm{a} + \mathrm{b} = 0,2 \\ \mathrm{và} 2\mathrm{a} + 4\mathrm{b} = 0,6 \\ \to \mathrm{a} = \mathrm{b} = 0,1 \\ \mathrm{Caot}: {\mathrm{n}}_{\mathrm{Cu}} = \frac{\mathrm{ne}}{2}=0,3 \mathrm{mol} \\ {\mathrm{n}}_{\mathrm{H}}+ = 4{\mathrm{n}}_{{\mathrm{O}}_2}\to {\mathrm{n}}_{\mathrm{NO}} =\frac{{\mathrm{n}}_{{\mathrm{H}}^{+}}}{4}=0,1 \mathrm{mol} \end{gathered}$$

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{A}\mathbf{ } \\ {\mathrm{n}}_{\mathrm{Y}} = {\mathrm{n}}_{\mathrm{X}} = 0,08 \mathrm{mol} \\ \mathrm{Các} \mathrm{axit} \mathrm{không} \mathrm{nhánh} \mathrm{nên} \mathrm{tối} \mathrm{đa} 2 \mathrm{chức}. \mathrm{Trong} \mathrm{Y} \mathrm{đặt} \mathrm{u}, \mathrm{v} \mathrm{là} \mathrm{số} \mathrm{mol} \mathrm{este} \mathrm{đơn} \mathrm{chức} \mathrm{và} \mathrm{hai} \mathrm{chức}. \\ {\mathrm{n}}_{\mathrm{Y}} = \mathrm{u} + \mathrm{v} = 0,08 \\ {\mathrm{n}}_{\mathrm{NaOH}} = \mathrm{u} + 2\mathrm{v} = 0,11 \\ \to \mathrm{u} = 0,05 \mathrm{và} \mathrm{v} = 0,03 \\ \mathrm{Đốt} 0,08 \mathrm{mol} \mathrm{Y} \mathrm{cần} {\mathrm{n}}_{{\mathrm{O}}_2}= 0,09.8 +\frac{0,17}{2}= 0,805 \mathrm{mol} \\ \mathrm{T} \mathrm{có} \mathrm{dạng} {\mathrm{C}}_{\mathrm{t}}{\mathrm{H}}_{2\mathrm{t}+2}\mathrm{O} (0,11 \mathrm{mol}) \\ \to {\mathrm{M}}_{\mathrm{T}} = 14\mathrm{t} + 18 = \frac{6,88}{0,11} \\ \to \mathrm{t} = \frac{35}{11} \\ \mathrm{Đốt} \mathrm{T} \mathrm{cần} {\mathrm{n}}_{{\mathrm{O}}_2}= 0,11.1,5\mathrm{t} = 0,525 \mathrm{mol} \\ \to \mathrm{Đốt} \mathrm{muối} \mathrm{cần} {\mathrm{n}}_{{\mathrm{O}}_2}= 0,805 – 0,525 = 0,28 \mathrm{mol} \\ \mathrm{Muối} \mathrm{gồm} {\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}-1}{\mathrm{O}}_2\mathrm{Na} (0,05 \mathrm{mol}) \mathrm{và} {\mathrm{C}}_{\mathrm{m}}{\mathrm{H}}_{2\mathrm{m}-4}{\mathrm{O}}_4{\mathrm{Na}}_2 (0,03 \mathrm{mol}) \\ \to {\mathrm{n}}_{{\mathrm{O}}_2} = \frac{0,05\left(3\mathrm{n}–2\right)}{2} + 0,03(1,5\mathrm{m} – 2,5) = 0,28 \\ \to 5\mathrm{n} + 3\mathrm{m} = 27 \\ \to \mathrm{n} = 3 \mathrm{và} \mathrm{m} = 4 \mathrm{là} \mathrm{nghiệm} \mathrm{duy} \mathrm{nhất}. \\ \mathrm{Vậy} \mathrm{muối} \mathrm{gồm} {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{COONa} (0,05) \mathrm{và} {\mathrm{C}}_2{\mathrm{H}}_4{\left(\mathrm{COONa}\right)}_2 (0,03) \\ \to \mathrm{\%}{\mathrm{m}}_{{\mathrm{C}}_2{\mathrm{H}}_4{\left(\mathrm{COONa}\right)}_2}= 50,31\% \end{gathered}$$