Đề thi thử THPT môn Hóa năm 2022 có đáp án (Mới nhất) - Đề 14

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{A} \\ \mathrm{X}:\text{}\left\{\begin{array}{l}\mathrm{COO}:0,03.3=0,09\\ {\mathrm{CH}}_2:\mathrm{a}\\ {\mathrm{H}}_2:\mathrm{b}\end{array}\right.+2,385{\text{ mol O}}_2\stackrel{{\mathrm{t}}^{\mathrm{o}}}{\to }{\mathrm{CO}}_2+1,57{\text{ mol H}}_2\mathrm{O} \\ \mathrm{Bảo} \mathrm{toàn} \mathrm{e}: 6\mathrm{a} + 2\mathrm{b} = 4. 2,385 \\ \mathrm{Bảo} \mathrm{toàn} \mathrm{H}: \mathrm{a} + \mathrm{b} = 1,57 \to \mathrm{a} = 1,6 \mathrm{mol}; \mathrm{b} = -0,03 \mathrm{mol} \\ \to 0,03 - {\mathrm{n}}_{\mathrm{\pi }\left(\mathrm{ankyl}\right) }= -0,03 \to {\mathrm{n}}_{\mathrm{\pi }\left(\mathrm{ankyl}\right)} = 0,06 \\ \to \mathrm{X} + 0,06 \mathrm{mol} {\mathrm{H}}_2 \to \mathrm{Y} ; \mathrm{Y} + 3\mathrm{KOH} \to \mathrm{muối} + {\mathrm{C}}_3{\mathrm{H}}_5{\left(\mathrm{OH}\right)}_3 {\mathrm{m}}_{\mathrm{X}}=0,09.44+1,6.14-0,03.2=26,3 \mathrm{gam} \\ \to 26,3 \mathrm{gam} \mathrm{X} \mathrm{có} \mathrm{số} \mathrm{mol} \mathrm{là} 0,03 \mathrm{mol}; \mathrm{phản} \mathrm{ứng} \mathrm{vừa} \mathrm{đủ} \mathrm{với} 0,06 \mathrm{mol} {\mathrm{H}}_2 \\ \to 65,75 \mathrm{gam} \mathrm{X} \mathrm{có} \mathrm{số} \mathrm{mol} \mathrm{là} 0,075 \mathrm{mol}; \mathrm{phản} \mathrm{ứng} \mathrm{vừa} \mathrm{đủ} \mathrm{với} 0,15 \mathrm{mol} {\mathrm{H}}_2 \mathrm{BTKL} \\ \to \mathrm{x} = 65,75 + 0,15.2 + 0,075.3.56 - 0,075.92 = 71,75\mathrm{g} \end{gathered}$$

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{A} \\ \left\{\begin{array}{l}\mathrm{X}:{\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{\mathrm{m}}\mathrm{N}\\ \mathrm{Y}:{\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{\mathrm{m}+1}{\mathrm{N}}_2\\ \mathrm{Anken}:{\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{2\mathrm{x}}\end{array}\right.\underset{}{\overset{+{\mathrm{O}}_2}{\to }}\left\{\begin{array}{l}{\mathrm{CO}}_2\text{ 0,22}\\ {\mathrm{H}}_2\mathrm{O}\text{ 0,3}\\ {\mathrm{N}}_2\text{ 0,03}\end{array}\right. \\ \mathrm{Từ} \mathrm{CTPT} \to \mathrm{X},\mathrm{Y} \mathrm{cùng} \mathrm{số} \mathrm{liên} \mathrm{kết} \\ \to {\text{n}}_{\mathrm{\pi }}=0,22+0,03-0,3+0,08=0,03<0,08 \to \mathrm{X}, \mathrm{Y} \mathrm{cùng} \mathrm{no} \\ {\mathrm{n}}_{\mathrm{anken}}={\text{n}}_{\mathrm{\pi }}=0,03\underset{}{\overset{}{\to }}{\mathrm{n}}_{\mathrm{X},\mathrm{Y}}=0,05\underset{}{\overset{}{\to }}\left\{\begin{array}{l}{\mathrm{n}}_{\mathrm{Y}}=0,03.2-0,05=0,01\\ {\mathrm{n}}_{\mathrm{X}}=0,04\end{array}\right. \\ \mathrm{BT}\left(\mathrm{C}\right):0,04\mathrm{n}+0,01\mathrm{n}+0,03\mathrm{x}=0,22\underset{}{\overset{}{\to }}\left\{\begin{array}{l}\mathrm{n}=3\\ \mathrm{x}=7/3\end{array}\right. \\ \underset{}{\overset{}{\to }}\mathrm{Y}:{\mathrm{C}}_3{\mathrm{H}}_{10}{\mathrm{N}}_2\text{ 0,01}\underset{}{\overset{}{\to }}\mathrm{\%}{\mathrm{m}}_{\mathrm{Y}}=18,14\mathrm{\%} \end{gathered}$$

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{A} \\ \mathrm{E}:\mathrm{HCOO}-{\mathrm{CH}}_2{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{OOCH}\text{ (o, m, p)} \\ \begin{array}{l}\mathrm{HCOO}-{\mathrm{CH}}_2{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{OOCH}+3\mathrm{NaOH}\underset{}{\overset{{\mathrm{t}}^{\mathrm{o}}}{\to }}2\mathrm{HCOONa}+{\mathrm{NaOC}}_6{\mathrm{H}}_4{\mathrm{CH}}_2\mathrm{OH}+{\mathrm{H}}_2\mathrm{O}\\ \text{ X Y}\end{array} \\ \to \mathrm{E}, \mathrm{X} \mathrm{tráng} \mathrm{bạc}, \mathrm{Y} \mathrm{tác} \mathrm{dụng} \mathrm{được} \mathrm{với} \mathrm{CO}2 \to \left(\mathrm{c}\right); \left(\mathrm{d}\right) \mathrm{đúng} \\ \begin{array}{l}2\mathrm{HCOONa}+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\underset{}{\overset{}{\to }}2\mathrm{HCOOH}+{\mathrm{Na}}_2{\mathrm{SO}}_4\\ \text{ X Z}\end{array} \\ \begin{array}{l}\text{HCOOH}+2{\mathrm{AgNO}}_3+\text{ 4}{\mathrm{NH}}_3+{\mathrm{H}}_2\mathrm{O}\underset{}{\overset{}{\to }} {\left({\mathrm{NH}}_4\right)}_2{\text{CO}}_3+2\mathrm{Ag}+2{\mathrm{NH}}_4{\mathrm{NO}}_3\\ \text{ Z T}\end{array} \\ \to \mathrm{T} \mathrm{là} \mathrm{hợp} \mathrm{chất} \mathrm{vô} \mathrm{cơ} \to \left(\mathrm{b}\right) \mathrm{sai} \\ \begin{array}{l}{\mathrm{NaOC}}_6{\mathrm{H}}_4{\mathrm{CH}}_2\mathrm{OH}+\mathrm{HCl}\underset{}{\overset{}{\to }}{\mathrm{HOC}}_6{\mathrm{H}}_4{\mathrm{CH}}_2\mathrm{OH}+\mathrm{NaCl}\\ \text{ Y F}\end{array} \\ \to \mathrm{F} \mathrm{là} \mathrm{hợp} \mathrm{chất} \mathrm{đa} \mathrm{chức} \to \left(\mathrm{e}\right) \mathrm{đúng}. \end{gathered}$$

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{C} \\ \{\begin{array}{l}\mathrm{X}\\ \mathrm{Y}\\ \mathrm{Z}\end{array}\underset{}{\overset{+\mathrm{NaOH}}{\to }}|\begin{array}{l}\text{F}\underset{}{\overset{+{\mathrm{O}}_2}{\to }}\left\{\begin{array}{l}{\mathrm{Na}}_2{\mathrm{CO}}_3\text{ 0,26}\\ {\mathrm{CO}}_2\\ {\mathrm{H}}_2\mathrm{O}\end{array}\right.\\ \mathrm{Ancol}\end{array} \\ \mathrm{Ancol}\underset{}{\overset{+\mathrm{Na}}{\to }} \mathrm{khối} \mathrm{lượng} \mathrm{bình} \mathrm{tăng} 16,2 \\ \to \{\begin{array}{l}{\text{n}}_{\mathrm{OH}\left(\mathrm{ancol}\right)}=0,26.2=0,52\\ {\mathrm{m}}_{\mathrm{ancol}}=16,2+0,52=16,72\end{array}\underset{}{\overset{}{\to }}0,26<{\mathrm{n}}_{\mathrm{ancol}}<0,52 \\ \underset{}{\overset{}{\to }}32,2<{\mathrm{M}}_{\mathrm{ancol}}<64,4 \end{gathered}$$

$$\begin{gathered} \to \mathrm{Ancol}\{\begin{array}{l}{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\text{ a}\\ {\mathrm{C}}_2{\mathrm{H}}_4{\left(\mathrm{OH}\right)}_2 \text{ b}\end{array}\underset{}{\overset{}{\to }}\{\begin{array}{l}\mathrm{a}+2\mathrm{b}=0,52\\ 46\mathrm{a}+62\mathrm{b}=16,72\end{array}\underset{}{\overset{}{\to }}\{\begin{array}{l}\mathrm{a}=0,04\\ \mathrm{b}=0,24\end{array} \\ \mathrm{BTKL}:{\mathrm{m}}_{\mathrm{F}}=38,56+0,52.40-16,72=42,64 \\ \text{F}\{\begin{array}{l}{\mathrm{R}}_1\mathrm{COONa}\text{ 0,26}\\ {\mathrm{R}}_2\mathrm{COONa}\text{ 0,26}\end{array}\underset{}{\overset{}{\to }}{\mathrm{m}}_{\mathrm{F}}=0,26.({\mathrm{R}}_1+{\mathrm{R}}_2)+0,52.67=42,64 \\ \underset{}{\overset{}{\to }}{\mathrm{R}}_1+{\mathrm{R}}_2=30\underset{}{\overset{}{\to }}\{\begin{array}{l}{\mathrm{R}}_1=1\\ {\mathrm{R}}_2=29\end{array} \\ \underset{}{\overset{}{\to }}\{\begin{array}{l}\mathrm{X}:{\mathrm{HCOOC}}_2{\mathrm{H}}_5\text{ 0,02}\underset{}{\overset{}{\to }}\mathrm{\%}{\mathrm{m}}_{{\mathrm{HCOOC}}_2{\mathrm{H}}_5}=3,84\mathrm{\%}\\ \mathrm{Y}:{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{COOC}}_2{\mathrm{H}}_5\text{ 0,02}\\ \mathrm{Z}:{\mathrm{HCOOCH}}_2-{\mathrm{CH}}_2-{\mathrm{OOCC}}_2{\mathrm{H}}_5\text{ 0,24}\end{array} \end{gathered}$$