[Năm 2022] Đề minh họa môn Hóa THPT Quốc gia có lời giải (Đề 17)

$\begin{array}{l}{C}_6{H}_{12}{O}_6+H_2\stackrel{}{\to }{C}_6{H}_{14}{O}_6\\ \mathrm{0,01}\leftarrow \mathrm{0,01}\\ \Rightarrow m_{glucozô}=\mathrm{0,01.180}:80\%=\mathrm{2,25}gam.\end{array}$

$+\left\{\begin{array}{l}{\pi }_X=3=2{\pi }_{chức}+1{\pi }_{gốcR}\\ \left(2\right)vàM_T=46\Rightarrow TlàC{H}_{3}OC{H}_3;ZlàC{H}_{3}OH\end{array}\right.\Rightarrow \left\{\begin{array}{l}Xlà\left[\begin{array}{l}C{H}_{3}OOCCH=CHCOOC{H}_3\\ C{H}_{3}OOCC(=C{H}_2)COOC{H}_3\end{array}\right.\\ YlàNaOOC{C}_2{H}_{2}COONa\end{array}\right.$

$\begin{array}{l}•X\text{\hspace{0.17em}\hspace{0.17em}kh}ôngphailaestecuaphenol\\ +\left\{\begin{array}{l}{n}_{X\left(C_x{H}_y{O}_2\right)}=n_{NaOH}=\mathrm{0,1}\\ 2\underset{\mathrm{0,1}}{\underbrace{n_{C_x{H}_y{O}_2}}}+2\underset{\mathrm{0,45}}{\underbrace{n_{O_2}}}=2\underset{\mathrm{0,4}}{\underbrace{n_{C{O}_2}}}+\underset{?}{\underbrace{n_{H_{2}O}}}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{H_{2}O}=\mathrm{0,3}\\ n_C:n_H:n_O=4:6:2\\ X\text{\hspace{0.17em}\hspace{0.17em}là\hspace{0.17em}\hspace{0.17em}}{C}_4{H}_6{O}_2\end{array}\right.\end{array}$

$\begin{array}{l}+X\mathit{ }có\mathit{ 3 }đồng\mathit{ }phân\mathit{ }cấu\mathit{ }tạo\mathit{ }có\mathit{ }dạng{C}_x{H}_{y}OOCH\text{\hspace{0.17em}\hspace{0.17em}là}:\hspace{0.17em}\\ \boxed{HCOOCH=CH-C{H}_3;HCOOC\left(C{H}_3\right)=C{H}_2;HCOOC{H}_{2}CH=C{H}_2}\\ •X là este của phenol\\ +\left\{\begin{array}{l}{n}_{X\left(C_x{H}_y{O}_2\right)}=\mathrm{0,5}{n}_{NaOH}=\mathrm{0,05}\\ 2\underset{\mathrm{0,05}}{\underbrace{n_{C_x{H}_y{O}_2}}}+2\underset{\mathrm{0,45}}{\underbrace{n_{O_2}}}=2\underset{\mathrm{0,4}}{\underbrace{n_{C{O}_2}}}+\underset{?}{\underbrace{n_{H_{2}O}}}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{H_{2}O}=\mathrm{0,2}\\ n_C:n_H:n_O=8:8:2\\ X\hspace{0.17em}\mathrm{là}{C}_8{H}_8{O}_2\end{array}\right.\\ +X có 3 đồng phân:\end{array}$

$\begin{array}{l}+n_{Ba{\left(OH\right)}_2}=\mathrm{0,3};n_{NaOH}=\mathrm{0,2};n_{BaC{l}_2}=\mathrm{0,24};n_{KOH}=\mathrm{0,3.}\\ +\underset{\mathrm{0,6}}{\underbrace{C{O}_2}}+\underset{ddX}{\underbrace{\left\{\begin{array}{l}Ba{\left(OH\right)}_2:\mathrm{0,3}\\ NaOH:\mathrm{0,2}\end{array}\right\}}}\Rightarrow 1<\frac{n_{O{H}^{-}}}{n_{C{O}_2}}=\frac{\mathrm{0,8}}{\mathrm{0,6}}<2\stackrel{}{\to }\left\{\begin{array}{l}C{O_3}^{2-}\\ HC{O_3}^{-}\end{array}\right\}\\ \Rightarrow n_{C{O_3}^{2-}}=n_{O{H}^{-}}-n_{C{O}_2}=\mathrm{0,2}<n_{B{a}^{2+}}\Rightarrow n_{BaC{O}_3}=\mathrm{0,2.}\\ \Rightarrow \underset{ddY}{\underbrace{\left\{\begin{array}{l}N{a}^{+}:\mathrm{0,2}\\ B{a}^{2+}:\mathrm{0,1}\\ HC{O_3}^{-}:\mathrm{0,4}\end{array}\right\}}}+\left\{\begin{array}{l}BaC{l}_2:\mathrm{0,24}\\ KOH:\mathrm{0,3}\end{array}\right\}\stackrel{}{\to }\left\{\begin{array}{l}{n}_{C{O_3}^{2-}}=\mathrm{0,3}<n_{B{a}^{2+}}=\mathrm{0,34}\\ n_{BaC{O}_3}=\mathrm{0,3}\iff \boxed{\mathrm{59,1}gam}\end{array}\right.\end{array}$

$$\begin{gathered} +\left\{\begin{array}{l}(\underset{6}{\underbrace{k}}-1)\underset{?}{\underbrace{n_{Xtrongmgam}}}=\underset{\mathrm{0,55}}{\underbrace{n_{C{O}_2}}}-\underset{\mathrm{0,5}}{\underbrace{n_{H_{2}O}}}\\ m_{Xtrongmgam}=856n_{Xtrongmgam}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{Xtrongmgam}=\mathrm{0,01}\\ m_{Xtrongmgam}=\mathrm{8,56}\end{array}\right. \\ +\left\{\begin{array}{l}\underset{\mathrm{0,06}}{\underbrace{n_{KOH}}}=3\underset{\mathrm{0,02}}{\underbrace{n_{C_3{H}_5{\left(OH\right)}_3}}}=3\underset{\mathrm{0,02}}{\underbrace{n_{Xtrong2mgam}}}\\ \underset{\mathrm{8,56.2}}{\underbrace{m_{Xtrong2mgam}}}+\underset{\mathrm{0,06.56}}{\underbrace{m_{KOH}}}=\underset{?}{\underbrace{m_{xàphòng}}}+\underset{\mathrm{0,02.92}}{\underbrace{m_{C_3{H}_5{\left(OH\right)}_3}}}\end{array}\right.\Rightarrow m_{xàphòng}=\boxed{\mathrm{18,64}gam} \end{gathered}$$

$\begin{array}{l}+\underset{\mathrm{39,36}gam}{\underbrace{\left\{\begin{array}{l}FeO\\ F{e}_3{O}_4\end{array}\right\}}}\underset{F{e}_3{O}_4=FeO+F{e}_2{O}_3}{\overset{quy\text{\hspace{0.17em}đổi}}{\to }}\underset{\mathrm{39,36}gam}{\underbrace{\left\{\begin{array}{l}FeO:xmol\\ F{e}_2{O}_3:ymol\end{array}\right\}}}\stackrel{H_{2}S{O}_{4}loãngdư}{\to }\left\{\begin{array}{l}FeS{O}_4:xmol\\ F{e}_2{\left(S{O}_4\right)}_3:ymol\\ H_{2}S{O}_4\end{array}\right\}\stackrel{KMn{O}_4}{\to }\left\{\begin{array}{l}MnS{O}_4\\ F{e}_2{\left(S{O}_4\right)}_3\\ K_{2}S{O}_4\end{array}\right\}\\ +\left\{\begin{array}{l}{m}_{hỗnhợp}=\mathrm{39,36}\\ BTE:n_{FeS{O}_4}=5n_{KMn{O}_4}\end{array}\right.\Rightarrow \left\{\begin{array}{l}72x+160y=\mathrm{39,36}\\ x=\mathrm{0,28}\end{array}\right.\Rightarrow \left\{\begin{array}{l}y=\mathrm{0,12}\\ BTE:n_{Cu}=n_{F{e}_2{\left(S{O}_4\right)}_3}=\mathrm{0,12}\end{array}\right.\Rightarrow \boxed{m_{Cu}=\mathrm{7,68}gam}\end{array}$

$\begin{array}{l}+\left\{\begin{array}{l}{\overline{C}}_{(X,Y)}=\frac{n_{C{O}_2}}{n_{(X,Y)}}=\mathrm{3,25}\\ {\overline{H}}_{(X,Y)}=\frac{2n_{H_{2}O}}{n_{(X,Y)}}=4\end{array}\right.\Rightarrow \left\{\begin{array}{l}Xlà\underset{xmol}{\underbrace{C_3{H}_4{O}_2}}(HCOOCH=C{H}_2)\\ Ylà\underset{ymol}{\underbrace{C_4{H}_4}}(CH\equiv C-CH=C{H}_2)\end{array}\right.\\ +\left\{\begin{array}{l}x+y=\mathrm{0,2}\\ 3x+4y=\mathrm{0,65}\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,15};y=\mathrm{0,05}\\ \%m_{C_4{H}_4}=\frac{\mathrm{0,05.52}}{\mathrm{0,05.52}+\mathrm{72.0,15}}.100\%=\boxed{\mathrm{19,40}\%}\end{array}\right.\end{array}$

$\begin{array}{l}+\left\{\begin{array}{l}{\overline{M}}_{Khí}=\mathrm{10,482}\\ 1khíhoánâu\end{array}\right.\Rightarrow khílà\left\{\begin{array}{l}NO\\ H_2\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{NO}+n_{H_2}=\mathrm{0,19}\\ 30n_{NO}+2n_{H_2}=\mathrm{0,19.10,842}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{NO}=\mathrm{0,06}\\ n_{H_2}=\mathrm{0,13}\end{array}\right.\\ +BTKL\Rightarrow n_{H_{2}O}=\frac{\mathrm{18,94}+\mathrm{0,43.98}-\mathrm{0,19.10,842}-\mathrm{54,34}}{18}=\mathrm{0,26}\Rightarrow n_{N{H_4}^{+}}=\frac{\mathrm{0,43.2}-\mathrm{0,13.2}-\mathrm{0,26.2}}{4}=\mathrm{0,02}\\ \Rightarrow n_{Cu{\left(N{O}_3\right)}_2}=\frac{\mathrm{0,06}+\mathrm{0,02}}{2}=\mathrm{0,04.}\\ +n_{H^{+}}=2n_{H_2}+4n_{NO}+10n_{N{H_4}^{+}}+2n_{O^{2-}}\Rightarrow n_{O^{2-}}=\frac{\mathrm{0,43.2}-\mathrm{2.0,13}-\mathrm{0,06.4}-\mathrm{0,02.10}}{2}=\mathrm{0,08}\Rightarrow n_{MgO}=\mathrm{0,08.}\\ +\left\{\begin{array}{l}27n_{Al}+65n_{Zn}=\mathrm{18,94}-\mathrm{0,04.188}-\mathrm{0,08.40}\\ 3n_{Al}+2n_{Zn}=\mathrm{0,06.3}+\mathrm{0,02.8}+\mathrm{0,13.2}\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,16}\\ y=\mathrm{0,06}\end{array}\right.\Rightarrow \boxed{\%Al=\mathrm{22,8}\%}\end{array}$

$+\left\{\begin{array}{l}Al\\ Ba\\ O\end{array}\right\}\stackrel{H_{2}O}{\to }Y\left\{\begin{array}{l}Ba{\left(OH\right)}_2:xmol\\ Ba{\left(Al{O}_2\right)}_2:ymol\end{array}\right\};\left\{\begin{array}{l}TN1:Y\stackrel{\mathrm{0,27}molC{O}_2}{\to }\left\{\begin{array}{l}Al{\left(OH\right)}_3\downarrow \\ BaC{O}_3\downarrow \end{array}\right\}+Ba{\left(HC{O}_3\right)}_2\\ TN1:Y\stackrel{C{O}_{2}dư}{\to } Al{\left(OH\right)}_3\downarrow +Ba{\left(HC{O}_3\right)}_2\end{array}\right.$

$\begin{array}{l}Các phản ứng tạo kết tủa:\left\{\begin{array}{l}Ba{\left(OH\right)}_2+C{O}_2\stackrel{}{\to }BaC{O}_3+H_{2}O\\ Ba{\left(Al{O}_2\right)}_2+2C{O}_2+4H_{2}O\stackrel{}{\to }Ba{\left(HC{O}_3\right)}_2+2Al\downarrow \end{array}\right.\\ \Rightarrow \left\{\begin{array}{l}TN1:197a+78.2y=\mathrm{21,51}\\ TN2:78.2y=\mathrm{15,6}\end{array}\right.\Rightarrow \left\{\begin{array}{l}a=\mathrm{0,03}\\ y=\mathrm{0,1}\end{array}\right.\Rightarrow n_{Ba}=\mathrm{0,03}+\frac{\mathrm{0,27}-\mathrm{0,03}}{2}=\mathrm{0,15}\\ \Rightarrow BTE:n_O=\frac{3n_{Al}+2n_{Ba}-2n_{H_2}}{2}=\frac{\mathrm{0,2.3}+\mathrm{0,15.2}-\mathrm{0,2.2}}{2}=\mathrm{0,25}mol\Rightarrow \boxed{m_X=\mathrm{29,95}gam}\end{array}$

$\begin{array}{l}+Các chất trong X đều có độ bất bão hòa k=2và có 2 nguyên tửO.\\ +\left\{\begin{array}{l}\underset{x}{\underbrace{n_{C{O}_2}}}-\underset{\mathrm{0,756}}{\underbrace{n_{H_{2}O}}}=n_{hỗn hợp}=\mathrm{0,5}\underset{y}{\underbrace{n_{OtrongX}}}\\ \underset{12x}{\underbrace{m_C}}+\underset{\mathrm{1,512}}{\underbrace{m_H}}+\underset{16y}{\underbrace{m_O}}=m_X=\mathrm{29,064}\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=\mathrm{1,176};y=\mathrm{0,084}\\ m_{BaC{O}_3}=\mathrm{1,176.197}=\boxed{\mathrm{231,672}gam}\end{array}\right.\end{array}$

$\begin{array}{l}+\left\{\begin{array}{l}Ankan\stackrel{chia nhỏ}{\to }C{H}_2+H_2\\ A\min no\stackrel{chia nhỏ}{\to }C{H}_2+NH+H_2\\ A\mathrm{mino}axitno\stackrel{chia nhỏ}{\to }COO+C{H}_2+NH+H_2\\ Chấtbéono\stackrel{chia nhỏ}{\to }COO+C{H}_2+H_2\end{array}\right\}\Rightarrow X\stackrel{chia nhỏ}{\to }\left\{\begin{array}{l}{H}_2:\mathrm{0,42}mol(=n_X)\\ NH:\mathrm{0,24}mol(=2n_{N_2})\\ C{H}_2:xmol\\ COO\end{array}\right\}\\ \Rightarrow \left\{\begin{array}{l}{n}_{H_{2}O}=\mathrm{0,42}+\mathrm{0,12}+x=\mathrm{3,32}\\ BTEchoX+O_2:4a=\mathrm{0,42.2}+\mathrm{0,24}+6x\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=\mathrm{2,78}\\ a=\mathrm{4,44}\end{array}\right.\end{array}$