[Năm 2022] Đề minh họa môn Hóa THPT Quốc gia có lời giải (Đề 16)

Đáp án A. Hg.

Đáp án D. Dầu hỏa.

$\begin{array}{l}{m}_{Fe}=\mathrm{2,52}\Rightarrow m_O=\mathrm{3,48}-\mathrm{2,52}=\mathrm{0,96}\Rightarrow n_O=\mathrm{0,06}\Rightarrow n_{A{l}_2{O}_3}=\mathrm{0,02}\\ \Rightarrow n_{Al}=\mathrm{0,04}\Rightarrow m_{Al}=\mathrm{1,08}gam.\end{array}$

$\left\{\begin{array}{l}FeS{O}_4\\ F{e}_2{\left(S{O}_4\right)}_3\end{array}\right.+NaOH\stackrel{}{\to }\left\{\begin{array}{l}Fe{\left(OH\right)}_2\\ Fe{\left(OH\right)}_3\end{array}\right.+HN{O}_3\stackrel{}{\to }Fe{\left(N{O}_3\right)}_3$

$\begin{array}{l}\left\{\begin{array}{l}Na\\ K\end{array}\right.+H_{2}O\stackrel{}{\to }\left|\begin{array}{l}\left\{\begin{array}{l}N{a}^{+}\\ O{H}^{-}\\ K^{+}\end{array}\right.+HCl\stackrel{}{\to }\left\{\begin{array}{l}N{a}^{+}\\ C{l}^{-}\\ K^{+}\end{array}\right.+H_{2}O\\ H_2:\mathrm{0,03}\end{array}\right.\\ n_{O{H}^{-}}=2.n_{H_2}=\mathrm{0,06}=n_{H^{+}}\Rightarrow V_{HCl}=\frac{\mathrm{0,06}}{\mathrm{0,1}}=\mathrm{0,6}lít=600ml\end{array}$

$\begin{array}{l}N{H}_2-C{H}_2-COOH+HCl\stackrel{}{\to }ClN{H}_3-C{H}_2-COOH\\ \mathrm{0,1}mol \to \mathrm{0,1}mol\\ \Rightarrow m_{muoi}=\mathrm{0,1.}\mathrm{111,5}=\mathrm{11,15}gam.\end{array}$

$$\begin{gathered} \underset{m\left(gam\right)}{\underbracket{hh\left\{\begin{array}{l}KHC{O}_3:\mathrm{0,8}m\\ CaC{O}_3:\mathrm{0,2}m\end{array}\right.}}\stackrel{t^o}{\to }ran\text{\hspace{0.17em}}Y\left\{\begin{array}{l}{K}_{2}C{O}_3\\ CaO\end{array}\right.\stackrel{H_{2}O}{\to }\left\{\begin{array}{l}ranZ:CaC{O}_3:\mathrm{0,2}m\\ ddE:\left\{\begin{array}{l}KOH\\ K_{2}C{O_3}_{d­}\end{array}\right.\stackrel{HCl1M}{\to }\left\{\begin{array}{l}Bat dau thoat khi:V_{1}lit\\ Khi thoat het:V_{2}lit\end{array}\right.\end{array}\right. \\ Vì dd E tác dụng với HCl tạo khí \Rightarrow K_{2}C{O}_3 dư \Rightarrow BT \left(Ca\right) \Rightarrow mCaC{O}_3 = 0,2m \left(gam\right) \Rightarrow m_{KHCO3} = 0,8m \left(gam\right) \end{gathered}$$

$Vì M_{CaCO3} = M_{KHCO3} \frac{n_{KHCO3}}{n_{CaCO3}}=\frac{\mathrm{0,8}m}{\mathrm{0,2}m}=4\Rightarrow Chon\left\{\begin{array}{l}{n}_{KHC{O}_3}=4mol\\ n_{CaC{O}_3}=1mol\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{K_{2}C{O}_3}=2mol\\ n_{CaO}=1mol\end{array}\right.$

$$\begin{gathered} CaO + H_{2}O \to Ca{\left(OH\right)}_2 \\ 1 mol \to 1 mol \\ Ca{\left(OH\right)}_2 + K_{2}C{O}_3 \to CaC{O}_3 + 2KOH \\ 1 mol < 2 mol \to 2 mol \\ Dư: 1 mol \end{gathered}$$

$$\begin{gathered} PTHH \left(1\right) O{H}^{-} + H^{+} \to H_{2}O \\ 2 2 \\ \left(2\right) C{O_3}^{2-} + H^{+} \to HC{O_3}^{-} \\ 1 1 1 \\ \left(3\right) HC{O_3}^{-} + H^{+} \to C{O}_2 + H_{2}O \\ 1 1 \\ \Rightarrow \frac{V_1}{V_2}=\frac{n_{H^{+}\left(\mathrm{1,2}\right)}}{n_{H^{+}\left(\mathrm{1,2,3}\right)}}=\frac{3}{4} \end{gathered}$$

$+\left\{\begin{array}{l}{n}_{muoi}=3n_{C_3{H}_5{\left(OH\right)}_3}=\mathrm{0,21}\\ n_{muoi}=\mathrm{2,5}x+\mathrm{1,75}x+x=\mathrm{0,21}\end{array}\right.\Rightarrow x=\mathrm{0,04}\Rightarrow X\stackrel{quy\text{\hspace{0.17em}\hspace{0.17em}do}i}{\to }\left\{\begin{array}{l}{C}_3{H}_5{\left(OH\right)}_3:\mathrm{0,07}\\ C_{15}{H}_{31}COOH:\mathrm{0,1}\\ C_{17}{H}_{33}COOH:\mathrm{0,07}\\ C_{17}{H}_{35}COOH:\mathrm{0,04}\\ H_{2}O:-\mathrm{0,21}mol\end{array}\right\}\Rightarrow m_X=\mathrm{59,36}gam$

$\begin{array}{l}+\mathrm{59,36}gamX\stackrel{O_2,t^o}{\to }BTE:n_{O_2}=\frac{\mathrm{14.0,07}+\mathrm{92.0,1}+\mathrm{102.0,07}+\mathrm{104.0,04}}{4}=\mathrm{5,37}mol\\ \Rightarrow \mathrm{47,488}gamX\stackrel{O_2,t^o}{\to }{n}_{O_2}=\frac{\mathrm{5,37.47,488}}{\mathrm{59,36}}=\boxed{\mathrm{4,296}mol}\end{array}$

$+\left\{\begin{array}{l}{n}_{O{H}^{-}}=n_{H^{+}\left(H_{2}O\right)}=2n_{H_2}=\mathrm{0,04}\\ \underset{\mathrm{0,04}}{\underbrace{n_{O{H}^{-}}}}=\underset{\mathrm{0,02}}{\underbrace{n_{H^{+}\left(HCl\right)}}}+2\underset{?}{\underbrace{n_{Cu{\left(OH\right)}_2}}}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{Cu{\left(OH\right)}_2}=\mathrm{0,01}mol\\ n_{Cu{\left(OH\right)}_2}=\boxed{\mathrm{0,98}gam}\end{array}\right.$

$\begin{array}{l}+\left\{\begin{array}{l}1<\frac{n_{KOH}}{n_X}<2\\ X+KOH\stackrel{}{\to }Y(no,don chuc, co phan ung trang bac\end{array}\right.\Rightarrow Xcó\left\{\begin{array}{l}estecủaphenol\\ estecủaankin\end{array}\right.\Rightarrow Ycódạng{C}_n{H}_{2n}O.\\ +\left\{\begin{array}{l}{n}_{\text{es}te\text{\hspace{0.17em}củ}aankin}+n_{\text{es}tecủaphenol}=\mathrm{0,1}\\ n_{\text{es}tecủaankin}+2n_{\text{es}tecủaphenol}=\mathrm{0,5}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{\text{es}tecủaankin}=\mathrm{0,1}\\ n_{\text{es}tecủaphenol}=\mathrm{0,2}\end{array}\right.\Rightarrow n_{C_n{H}_{2n}O}=\mathrm{0,1}mol.\\ +BTEkhiYcháy:\mathrm{0,1.}(6n-2)=\mathrm{0,25.4}\Rightarrow n=2\Rightarrow YlàC{H}_{3}CHO.\\ +Tacó:X+\underset{\mathrm{0,5}mol}{\underbrace{KOH}}\stackrel{}{\to }\underset{53gam}{\underbrace{muoi}}+\underset{\mathrm{0,1}mol}{\underbrace{C{H}_{3}CHO}}+\underset{\mathrm{0,2}mol}{\underbrace{H_{2}O}}\Rightarrow \boxed{m_X=33gam}\end{array}$

$\begin{array}{l}\left\{\begin{array}{l}F{e}_x{O}_y\\ Cu,CuO\end{array}\right.\stackrel{+\mathrm{0,3}molHCl}{\to }\left\{\begin{array}{l}F{e}^{2+}\\ C{u}^{2+}\\ C{l}^{-}:\mathrm{0,3}\end{array}\right.\stackrel{+AgN{O}_3}{\to }\left\{\begin{array}{l}AgCl:\mathrm{0,3}mol\\ Ag:\frac{\mathrm{51,15}-\mathrm{0,3.143,5}}{108}=\mathrm{0,075}mol\end{array}\right.\\ \Rightarrow n_{F{e}^{2+}}=n_{A{g}^{+}phan ung}=n_{Ag}=\mathrm{0,075}mol\Rightarrow n_{C{u}^{2+}}=(\mathrm{0,3}-\mathrm{0,075.2}):2=\mathrm{0,075}mol.\\ +2.n_{O^{2-}}=n_{C{l}^{-}}=\mathrm{0,3}mol\Rightarrow n_{O^{2-}}=\mathrm{0,15}mol.\\ \Rightarrow m_X=\mathrm{3,2}+\mathrm{0,075.56}+\mathrm{0,075.64}+\mathrm{0,15.16}=\mathrm{15,6}gan nhat voi\mathrm{14,5.}\end{array}$

$\begin{array}{l}+\mathrm{0,2}molX\stackrel{quy\text{\hspace{0.17em}\hspace{0.17em}do}i}{\to }\left\{\begin{array}{l}{C}_x{H}_y{N}_t:\mathrm{0,2}mol\\ C{O}_2\end{array}\right\}\stackrel{O_2}{\to }\left\{\begin{array}{l}C{O}_2\\ H_{2}O:\mathrm{0,48}mol\\ N_2:\mathrm{0,07}mol\end{array}\right\}\Rightarrow \left\{\begin{array}{l}BTH:\mathrm{0,2}y=\mathrm{0,48.2}\\ BTN:\mathrm{0,2}t=\mathrm{0,07.2}\end{array}\right.\Rightarrow \left\{\begin{array}{l}y=\mathrm{4,8}\\ t=\mathrm{0,7}\end{array}\right.\\ +\mathrm{0,2}molX+\mathrm{0,07}molB{r}_2\Rightarrow \mathrm{0,2}{k}_X=\mathrm{0,07}\Rightarrow \mathrm{0,2.}\frac{2x-y+t+2}{2}=\mathrm{0,07}\Rightarrow x=\mathrm{1,4}\\ \Rightarrow n_{O_2}=n_C+\frac{n_{H_{2}O}}{2}=\mathrm{1,4.0,2}+\frac{\mathrm{0,48}}{2}=\mathrm{0,52}gan nhat voi\boxed{\mathrm{0,5}}\end{array}$

$\begin{array}{l}+E\stackrel{quy\text{\hspace{0.17em}\hspace{0.17em}do}i}{\to }\left\{\begin{array}{l}H\\ C\\ COO\end{array}\right\}\Rightarrow \left\{\begin{array}{l}BTH:n_H=2n_{H_{2}O}=\mathrm{0,55}\\ BTE:n_C=\frac{4n_{O_2}-n_H}{4}=\mathrm{0,26}\end{array}\right.\Rightarrow \left\{\begin{array}{l}BTKL:n_{COO}=\frac{\mathrm{6,75}-\mathrm{0,26.12}-\mathrm{0,55}}{44}=\mathrm{0,07}\\ n_{2ancol}=n_{2muoiX,Y}=\mathrm{0,07}mol\end{array}\right.\end{array}$

$\begin{array}{l}\\ +2ancol\stackrel{quy\text{\hspace{0.17em}\hspace{0.17em}do}i}{\to }\left\{\begin{array}{l}C{H}_{3}OH:\mathrm{0,07}mol\\ C{H}_2:ymol\end{array}\right\}\Rightarrow BTE:\mathrm{0,07.6}+6y=\mathrm{0,18.4}\Rightarrow y=\mathrm{0,05}\iff \left\{\begin{array}{l}C{H}_{3}C{H}_{2}OH:\mathrm{0,05}\\ C{H}_{3}OH:\mathrm{0,02}\end{array}\right\}\end{array}$

$\Rightarrow \left\{\begin{array}{l}{n}_{CtrongX,Y}=\mathrm{0,26}-\mathrm{0,05.2}-\mathrm{0,02}=\mathrm{0,14}\\ n_{HtrongX,Y}=\mathrm{0,55}-\mathrm{0,05.5}-\mathrm{0,02.3}=\mathrm{0,24}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{C}_{(X,Y)}=\mathrm{0,14}:\mathrm{0,07}=2\\ H_{(X,Y)}=\mathrm{0,24}:\mathrm{0,07}=\mathrm{3,42}\end{array}\right.$

$\Rightarrow \left\{\begin{array}{l}XlàC{H}_3-C{H}_{2}COONa:\mathrm{0,015}\\ YlàC{H}_2=CHCOONa:\mathrm{0,055}\end{array}\right.\Rightarrow thoa man hoac\left\{\begin{array}{l}XlàC{H}_3-C{H}_{2}COONa:\mathrm{0,0425}mol\\ YlàCH\equiv CCOONa:\mathrm{0,0275}mol\end{array}\right.\Rightarrow loai.$