C1: B

C2: A

C3: C

C4: B

C5: B

C6: B

PT: \(2R+O_2\underrightarrow{t^o}2RO\)

Ta có: \(n_R=\dfrac{9,6}{M_R}\left(mol\right)\)

\(n_{RO}=\dfrac{16}{M_R+16}\left(mol\right)\)

Theo PT: \(n_R=n_{RO}\Rightarrow\dfrac{9,6}{M_R}=\dfrac{16}{M_R+16}\Rightarrow M_R=24\left(g/mol\right)\)

→ R là Mg.

C8: Ta có: \(n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\)

PT: \(2K+2H_2O\rightarrow2KOH+H_2\)

____0,1____________0,1____0,05 (mol)

⇒ m_KOH = 0,1.56 = 5,6 (g)

V_H2 = 0,05.22,4 = 1,12 (l)

C9: A

PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

Theo PT: \(n_{O_2}=2n_{Fe_3O_4}=0,4\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,4.22,4=8,96\left(l\right)\)

C10: A

PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

- Khí bay lên là O₂.

Ta có: \(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,4\left(mol\right)\)

\(\Rightarrow m_{KMnO_4}=0,4.158=63,2\left(g\right)\)

\(PTHH:2KMnO_4-^{t^o}>K_2MnO_4+MnO_2+O_2\)

n(mol)          0,4<----------------------------------------0,2

=> khí bay lên là oxi

\(n_{O_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(m_{KMnO_4}=n\cdot M=0,4\cdot\left(39+55+16\cdot4\right)=63,2\left(g\right)\)

Ta có: \(n_{CO_2}=\dfrac{4,4}{44}=0,1\left(mol\right)\)

\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(\Rightarrow\overline{M}_B=\dfrac{m_{CO_2}+m_{O_2}+m_{H_2}}{n_{CO_2}+n_{O_2}+n_{H_2}}=\dfrac{4,4+0,1.32+0,3.2}{0,1+0,1+0,3}=16,4\left(g/mol\right)\)

\(\Rightarrow d_{B/H_2}=\dfrac{\overline{M}_B}{M_{H_2}}=\dfrac{16,4}{2}=8,2\)

a, PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

Ta có: \(n_{Al}=\dfrac{10,2}{27}=\dfrac{17}{45}\left(mol\right)\)

b, Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=\dfrac{17}{60}\left(mol\right)\)

\(\Rightarrow V_{O_2}=\dfrac{17}{60}.22,4\approx6,347\left(l\right)\)

c, Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=\dfrac{17}{90}\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3}=\dfrac{17}{90}.102\approx19,267\left(g\right)\)

d, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

Theo PT: \(n_{KMnO_4}=2n_{O_2}=\dfrac{17}{30}\left(mol\right)\)

\(\Rightarrow m_{KMnO_3}=\dfrac{17}{30}.158\approx89,53\left(g\right)\)

a, Không pư

b, \(CO_2+2KOH\rightarrow K_2CO_3+H_2O\)

c, Không pư.

d, \(CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O\)

Buta-1,3-đien : $CH_2=CH-CH=CH_2$

$CH_2=CH-CH=CH_2 + 2H_2 \xrightarrow{t^o,Ni} CH_3-CH_2-CH_2-CH_3$

$C_2H_4 + Br_2 \to C_2H_4Br_2$
Theo PTHH : 

$n_{C_2H_4} = n_{Br_2} = 0,1.2 = 0,2(mol)$
$\%m_{C_2H_4} = \dfrac{0,2.28}{10}.100\% = 56\%$

$\%m_{CH_4} = 100\% - 56\% = 44\%$

\(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\\ PTHH:Fe+2HCl->FeCl_2+H_2\)

ti le        1      :    2       :      1      :    1

n(mol)     0,5-->1--------->0,5------>0,5

\(m_{FeCl_2}=n\cdot M=0,5\cdot\left(56+35,5\cdot2\right)=63,5\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,5\cdot22,4=11,2\left(l\right)\)

\(n_{O_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ PTHH:2KClO_3-^{t^o}>2KCl+3O_2\)

ti le            2         :         2        :      3

n(mol)     0,2<---------------0,2<------0,3

\(m_{KClO_3}=n\cdot M=0,2\cdot\left(39+35,5+16\cdot3\right)=24,5\left(g\right)\\ m_{KCl}=n\cdot M=0,2\cdot\left(39+35,5\right)=14,9\left(g\right)\)

- Hòa tan hỗn hợp vào nước thu được ddA: (CH₃COO)₂Mg, (CH₃COO)₂Ba, CH₃COONa 

- Sục khí NH₃ dư vào ddA, lọc tách thu được Mg(OH)₂ và ddB: CH₃COONH₄, NH₃, (CH₃COO)₂Ba, CH₃COONa.

\(\left(CH_3COO\right)_2Mg+2NH_3+2H_2O\rightarrow Mg\left(OH\right)_2\downarrow+2CH_3COONH_4\)

- Hòa tan Mg(OH)₂ và CH₃COOH dư, cô cạn dd sau phản ứng thu được (CH₃COO)₂Mg khan

\(Mg\left(OH\right)_2+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+2H_2O\)

- Cho dd (NH₄)₂CO₃ dư vào ddB, lọc tách thu được BaCO₃ và ddC: (NH₄)₂CO₃, CH₃COONH₄, CH₃COONa, NH₃

\(\left(NH_4\right)_2CO_3+\left(CH_3COO\right)_2Ba\rightarrow BaCO_3\downarrow+2CH_3COONH_4\)

- Hòa tan BaCO₃ vào dd CH₃COOH dư, cô cạn dd sau phản ứng thu được (CH₃COO)₂Ba khan

\(BaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ba+CO_2\uparrow+H_2O\)

- Cô cạn ddC thu được CH₃COONa khan

\(CH_3COONH_4\xrightarrow[]{t^o}CH_3COOH+NH_3\uparrow\\ \left(NH_4\right)_2CO_3\xrightarrow[]{t^o}2NH_3\uparrow+CO_2\uparrow+H_2O\)