[Năm 2022] Đề minh họa môn Hóa THPT Quốc gia có lời giải (Đề 4)

Chọn đáp án A.C₁₇H₃₅COOH.

Chọn đáp án A. Etylamin.

Chọn đáp án A. Lysin.

Chọn đáp án C. Poli(metyl metacrylat).

Chọn đáp án A. Ag.

Chọn đáp án D. Cu.

Chọn đáp án A. Cu.

Chọn đáp án B. Na.

Chọn đáp án A. O₂.

Chọn đáp án B. Etilen.

Chọn đáp án D. 1.

Saccrarozơ + H₂O $\stackrel{H^{+}}{\to }$Glucozơ + Fructozơ

Glucozơ + H₂ $\stackrel{Ni,t^o}{\to }$ Sobitol

Fructozơ + H₂ $\stackrel{Ni,t^o}{\to }$ Sobitol

$C_6{H}_{10}{O}_5\stackrel{}{\to }{C}_6{H}_{12}{O}_6\stackrel{}{\to }2Ag$

$\begin{array}{l}BTNT\left(N\right):n_{HCl}=n_{N\left(X\right)}=2n_{N_2}=\mathrm{0,2}mol\\ \stackrel{}{\to }{n}_{N_2}=\mathrm{0,1}mol\stackrel{}{\to }{V}_{N_2}=\mathrm{2,24}\left(l\right)\end{array}$

Chọn đáp án C. Tơ nitron được điều chế bằng phản ứng trùng ngưng.

$\left\{\begin{array}{l}\left[m\right]:27.n_{Al}+56.n_{Fe}=\mathrm{13,8}\\ \left[BTE\right]:3.n_{Al}+2.n_{Fe}=\mathrm{2.0,45}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{Al}=\mathrm{0,2}\\ n_{Fe}=\mathrm{0,15}\end{array}\right.\Rightarrow \%m_{Al/X}=\mathrm{39,13}\%$

Ta có: $\left\{\begin{array}{l}{n}_{glyxerol}=\underset{x}{\underbrace{n_X}}\Rightarrow n_{glyxerol}=x\\ n_{NaOH}=3\underset{x}{\underbrace{n_X}}\Rightarrow n_{NaOH}=3x\end{array}\right.\stackrel{BTKL}{\to }\mathrm{158,4}+3x.40=\mathrm{163,44}+x.92\Rightarrow x=\mathrm{0,18}$

Đặt CT chung của X là C_nH_2nO₂

=> Đốt cháy X tạo n_CO2 = n_H2O

Mà m_CO2 + m_H2O = 40,3

=> n_CO2 = n_H2O =$\frac{\mathrm{40,3}}{44+18}$= 0,65 mol

nNaOH = 0,2 mol => nX = 0,2 => nO(trong X) = 0,4 mol

Bảo toàn khối lượng => mX = 0,4.16 + 0,65.(12+2) = 15,5

Mà X + $\frac{V}{\mathrm{22,4}}$mol O₂ -> CO₂ + H₂O

Bảo toàn khối lượng => 15,5 +$\frac{V}{\mathrm{22,4}}$.32 = 40,3 => V = 17,36

$\begin{array}{l}+\left\{\begin{array}{l}t.n_{C_{\overline{n}}{H}_{2\overline{n}+2+t}{N}_t}=2n_{N_2}=\mathrm{0,18}\\ (\underset{0}{\underbrace{k}}-1-\mathrm{0,5}\underset{?=}{\underbrace{t}})n_{C_{\overline{n}}{H}_{2\overline{n}+2+t}{N}_t}=\underset{\mathrm{0,75}}{\underbrace{n_{C{O}_2}}}-\underset{\mathrm{0,93}}{\underbrace{n_{H_{2}O}}}\end{array}\right.\Rightarrow \left\{\begin{array}{l}t.n_{C_{\overline{n}}{H}_{2\overline{n}+2+t}{N}_t}=\mathrm{0,18}\\ n_{C_{\overline{n}}{H}_{2\overline{n}+2+t}{N}_t}=\mathrm{0,09}\end{array}\right.\\ \Rightarrow \left\{\begin{array}{l}t=2\\ BTC:\mathrm{0,09}\overline{n}+3.(\mathrm{0,3}-\mathrm{0,09})=\mathrm{0,75}\end{array}\right.\Rightarrow \left\{\begin{array}{l}\overline{n}=\frac{4}{3}\\ haia\min \text{\hspace{0.17em}\hspace{0.17em}là\hspace{0.17em}\hspace{0.17em}}C{H}_6{N}_2\text{\hspace{0.17em}\hspace{0.17em}và\hspace{0.17em}\hspace{0.17em}}{C}_2{H}_8{N}_2\end{array}\right.\\ +\left\{\begin{array}{l}{n}_{C{H}_6{N}_2}+n_{C_2{H}_8{N}_2}=\mathrm{0,09}\\ {\overline{C}}_{2a\min }=\frac{n_{C{H}_6{N}_2}+2_{C_2{H}_8{N}_2}}{\mathrm{0,09}}=\frac{4}{3}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{C{H}_6{N}_2}=\mathrm{0,06};n_{C_2{H}_8{N}_2}=\mathrm{0,03}\\ m_{C{H}_6{N}_2}=\mathrm{0,06.46}=\boxed{\mathrm{2,76}gam}\end{array}\right.\end{array}$

$\underset{\mathrm{0,15}\text{\hspace{0.33em}mol}}{\underbrace{C{O}_2}}+\left\{\begin{array}{l}\underset{\mathrm{0,1}\text{ mol}}{\underbrace{KOH}}\\ \underset{\mathrm{0,16}\text{\hspace{0.33em}mol}}{\underbrace{{\text{K}}_{2}C{O}_3}}\end{array}\right.\stackrel{}{\to }\left|\begin{array}{l}\left\{\begin{array}{l}\underset{a\text{ mol}}{\underbrace{K_{2}C{O}_3}}\\ \underset{b\text{ mol}}{\underbrace{{\text{KHCO}}_3}}\end{array}\right.\stackrel{BaC{l}_2\text{\hspace{0.33em}d­}}{\to }\left|\begin{array}{l}\underset{a\text{\hspace{0.33em}mol}}{\underbrace{BaC{O}_3\downarrow }}\stackrel{t^0}{\to }\left|\begin{array}{l}\underset{a\text{ mol}}{\underbrace{BaO}}:m\text{ gam}\\ {\text{CO}}_2\end{array}\right.\\ \left\{\begin{array}{l}KCl\\ KHC{O}_3\end{array}\right.\end{array}\right.\\ H_{2}O\end{array}\right.$

$\begin{array}{l}\text{Thí nghiệm 1:}\left\{\begin{array}{l}\stackrel{BTNT.C}{\to }a+b=\mathrm{0,15}+\mathrm{0,16}\\ \stackrel{BTNT.K}{\to }2a+b=\mathrm{0,1}+\mathrm{2.0,16}\end{array}\right.\to \left\{\begin{array}{l}a=\mathrm{0,11}\\ b=\mathrm{0,2}\end{array}\right.\\ \to n_{BaO}\stackrel{BTNT.Ba}{=}{n}_{BaC{O}_3}\stackrel{BTNT.C\text{ cho t}\mathrm{hí} \mathrm{nghiệm}\text{ 2}}{=}\mathrm{0,11}\text{ mol}\to \text{m = 153.0,11=16,83\hspace{0.33em}gam}\end{array}$

$\begin{array}{l}pH=13\to pOH=1\to n_{O{H}^{-}d­}={10}^{-1}\mathrm{.0,4}=\mathrm{0,04}\text{mol}\\ n_{O{H}^{-}\text{\hspace{0.17em}p­}}=n_{H^{+}}=n_{HCl}+2n_{H_{2}S{O}_4}=\mathrm{0,04}+\mathrm{2.0,03}=\mathrm{0,1}\text{\hspace{0.33em}mol }\\ \to {\text{n}}_{O{H}^{-}\left(200mlY\right)}=\mathrm{0,14}\text{\hspace{0.33em}mol}\to {\text{n}}_{O{H}^{-}\left(400mlY\right)}=\mathrm{0,28}\text{\hspace{0.33em}mol}\end{array}$

$\begin{array}{l}\stackrel{BTNT.H}{\to }2n_{H_{2}O}=n_{O{H}^{-}(400ml\text{\hspace{0.33em}Y)}}+2n_{H_2}\to n_{H_{2}O}=\mathrm{0,21}\\ \stackrel{BTNT.O}{\to }{n}_{O\left(X\right)}=n_{O{H}^{-}(400ml\text{\hspace{0.33em}Y)}}-n_{H_{2}O}=\mathrm{0,07}\\ \to {\text{m}}_X=\frac{100}{\mathrm{8,75}}{m}_O=\frac{100}{\mathrm{8,75}}\mathrm{.16.0,07}=\mathrm{12,8}\text{\hspace{0.33em}gam}\stackrel{\text{gần nhất}}{\to }\text{ B}\end{array}$

Các phản ứng xảy ra khi tiến hành các thí nghiệm (với điều kiện tương ứng) là:

☒ (a) 4HCl + NaAlO₂ → NaCl + AlCl₃ + 2H₂O.

☑ (b) CO₂ + NaAlO₂ + H₂O → Al(OH)₃↓ + NaHCO₃.

☑ (c) 4Ba(OH)₂ + Al₂(SO₄)₃ → Ba(AlO₂)₂ + 3BaSO4↓ + 4H₂O.

☑ (d) 6NH₃ + Al₂(SO₄)₃ + 6H₂O → 2Al(OH)₃↓ + 3(NH₄)₂SO₄.

☑ (e) AlCl₃ + 3NaOH → Al(OH)₃↓ + 3NaCl.

* $\underset{\text{este}}{\underbrace{\text{E}}}\left\{\begin{array}{l}\text{X}\\ \text{Y}\\ \text{Z}\end{array}\right.{\text{(M}}_{\text{X}}{\text{<M}}_{\text{Y}}{\text{<M}}_{\text{Z}}\text{)}\stackrel{\text{NaOH}}{\to }\left\{\begin{array}{l}\text{Ancol T}\stackrel{\text{Na}}{\to }{\text{0,2 mol H}}_{\text{2}}{\text{+m}}_{\text{b}\uparrow }\text{=12 gam}\\ \text{F}\left\{\begin{array}{l}\text{G: 5x}\\ \text{H: 3x}\end{array}\right.\text{(mol)}\stackrel{{\text{+O}}_{\text{2}}}{\to }{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{+CO}}_{\text{2}}{\text{+0,35 mol H}}_{\text{2}}\text{O}\end{array}\right.$

* ${\text{n}}_{\text{NaOH}}{\text{= n}}_{\text{OH(ancol)}}{\text{= 2n}}_{{\text{H}}_{\text{2}}}\text{= 0,4(mol)}$ $\Rightarrow$ ${\text{M}}_{\text{Ancol}}\text{=}\frac{(12+\mathrm{0,2.2})}{\frac{\text{0,4}}{\text{x}}}\text{=31x}$ $\Rightarrow$Ancol là C₂H₄(OH)₂

Chọn đáp án  A. 150,32.

Trong không khí, một khí không màu hóa nâu trong không khí là NO,

$$\begin{gathered} {\overline{\text{M}}}_Z=\mathrm{10,8.2}=\mathrm{21,6}<M_{NO}\to M_{\text{khí còn lại }}<\mathrm{21,6}\to khí còn lại là {\text{H}}_2 \\ Z\left\{\begin{array}{l}NO:a\text{\hspace{0.33em}mol}\\ H_2:b\text{\hspace{0.33em}mol}\end{array}\right.\to \left\{\begin{array}{l}a+b=\frac{\mathrm{2,24}}{\mathrm{22,4}}=\mathrm{0,1}\text{\hspace{0.33em}mol}\\ \text{30a + 2b = 0,1.21,6}\end{array}\right.\to \left\{\begin{array}{l}a=\mathrm{0,07}\\ b=\mathrm{0,03}\end{array}\right. \end{gathered}$$

$X\left\{\begin{array}{l}Mg:x\text{\hspace{0.33em}mol}\\ {\text{Fe}}_3{O}_4:y\text{\hspace{0.33em}mol}\\ {\text{Fe(NO}}_3^{}{)}_2:z\text{\hspace{0.33em}mol}\end{array}\right.\to 24x+232y+180z=\mathrm{17,32}\text{\hspace{0.33em}gam (1)}$

$\begin{array}{l}\stackrel{BTNT.N}{\to }{n}_{N{H}_4^{+}\left(Y\right)}=n_{HN{O}_3}+2n_{Fe{\left(N{O}_3\right)}_2}-n_{NO}=\mathrm{0,08}+2z-\mathrm{0,07}=2z+\mathrm{0,01}\\ {\text{n}}_{H^{+}}=2n_{O\left(F{e}_3{O}_4\right)}+4n_{NO}+10n_{N{H}_4^{+}}+2n_{H_2}\\ \to (\mathrm{1,04}+\mathrm{0,08})=8y+\mathrm{4.0,07}+10(2z+\mathrm{0,01})+\mathrm{2.0,03}=\text{\hspace{0.33em} (2)}\end{array}$

$\text{20,8\hspace{0.33em}gam chất rắn }\left\{\begin{array}{l}MgO\stackrel{BTNT.Mg}{\to }{n}_{MgO}=x\text{\hspace{0.33em}mol}\\ F{e}_2{O}_3\stackrel{BTNT.Fe}{\to }{n}_{F{e}_2{O}_3}=(\mathrm{1,5}y+\mathrm{0,5}z)mol\end{array}\right.$

$\text{Từ (1), (2) và (3) }\to \left\{\begin{array}{l}x=\mathrm{0,4}\\ y=\mathrm{0,01}\\ z=\mathrm{0,03}\end{array}\right.\to Y\left\{\begin{array}{l}M{g}^{2+}:\mathrm{0,4}\text{\hspace{0.33em}mol}\\ {\text{Fe}}^{2+}:a\text{\hspace{0.33em}mol}\\ {\text{Fe}}^{3+}:b\text{\hspace{0.33em}mol}\\ {\text{NH}}_4^{+}:\mathrm{0,07}\text{\hspace{0.33em}mol}\\ {\text{Cl}}^{-}:\mathrm{1,04}\text{\hspace{0.33em}mol}\end{array}\right.$

$\to \left\{\begin{array}{l}\stackrel{BTNT.Fe}{\to }a+b=\mathrm{3.0,01}+\mathrm{0,03}\\ \stackrel{\text{BTĐT}}{\to }2a+3b+\mathrm{2.0,4}+\mathrm{0,07}=\mathrm{1,04}\end{array}\right.\to \left\{\begin{array}{l}a=\mathrm{0,01}\\ b=\mathrm{0,05}\end{array}\right.$

$\to m\text{\hspace{0.33em}gam kết tủa }\left\{\begin{array}{l}Ag\stackrel{BTE}{\to }{n}_{Ag}=n_{F{e}^{2+}}=\mathrm{0,01}\text{\hspace{0.33em}mol}\\ AgCl\stackrel{BTNT.Cl}{\to }{n}_{AgCl}=n_{C{l}^{-}\left(Y\right)}=\mathrm{1,04}\text{\hspace{0.33em}mol}\end{array}\right.$

$\to m=\mathrm{108.0,01}+\mathrm{143,5.1,04}=\mathrm{150,32}\text{\hspace{0.33em}gam -> Đ}áp á\text{n A}$