\(2CO + O_2 \xrightarrow{t^o} 2CO_2\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ n_{CO} = n_{CO_2} = \dfrac{8,8}{44} = 0,2(mol)\\ n_{O_2} = \dfrac{n_{CO} + n_{H_2}}{2}=\dfrac{0,2+n_{H_2}}{2} = \dfrac{9,6}{32} = 0,3(mol)\\ \Rightarrow n_{H_2} = 0,4(mol)\\ \%V_{CO} = \dfrac{0,2}{0,2 + 0,4}.100\% = 33,33\%\\ \%V_{H_2} = 100\% - 33,33\% = 66,67\%\\ \%m_{CO} = \dfrac{0,2.28}{0,2.28+0,4.2}.100\%=87,5\%\\ \%m_{H_2} = 100\% - 87,5\% = 12,5\%\)

nCO2 = 8.8/44 =  0.2 (mol) 

nO2 = 9.6/32 = 0.3 (mol) 

2CO + O2 -to-> 2CO2 

0.2____0.1______0.2

2H2 + O2 -to-> 2H2O 

0.4___0.3-0.1

%CO = 0.2*28 / ( 0.2*28 + 0.4*2) * 100% = 87.5%

%H2 = 12.5% 

=> D

\(2CO + O_2\xrightarrow{t^o} 2CO_2(1)\\ n_{CO} = n_{CO_2} = \dfrac{8,8}{44} = 0,2(mol)\\ n_{O_2(1)} = \dfrac{n_{CO_2}}{2} = 0,1(mol)\\ 2H_2 +O_2 \xrightarrow{t^o} 2H_2O(2)\\ n_{H_2} = 2n_{O_2(2)} = 2.(\dfrac{9,6}{32}-0,1) = 0,4(mol)\\ \Rightarrow \%m_{CO} = \dfrac{0,2.28}{0,2.28 + 0,4.2}.100\% = 87,5\%\\ \%m_{H_2} = 100\% - 87,5\% = 12,5\%\)

PTHH: \(2CO+O_2\underrightarrow{t^o}2CO_2\)  (1)

            \(2H_2+O_2\underrightarrow{t^o}2H_2O\)  (2)

Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\\n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{CO}=0,2\left(mol\right)\\n_{H_2}=0,4\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CO}=0,2\cdot28=5,6\left(g\right)\\m_{H_2}=0,4\cdot2=0,8\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{CO}=\dfrac{5,6}{5,6+0,8}\cdot100\%=87,5\%\\\%m_{H_2}=12,5\%\end{matrix}\right.\)

PTHH: 2CO+O2to→2CO2 (1)

                4H2+O2to→2H2O  (2)

b) Ta có:

ΣnO2=\(\dfrac{9,6}{32}\)=0,3(mol)

nCO2=\(\dfrac{8,8}{44}\)=0,2(mol)

⇒{nO2(1)=0,1mol

nO2(2)=0,2mol

⇒{mCO=0,1⋅28=2,8(g)

mH2=0,2⋅2=0,4(g)

 ⇒%mCO=\(\dfrac{2,8}{2,8+0,4}\)⋅100%=87,5%

%mH2=12,5%

\(nO_2=\dfrac{9,6}{32}=0,3\left(mol\right)\)

\(nCO_2=\dfrac{8,8}{44}=0,2\left(mol\right)\)

\(2CO+O_2\underrightarrow{t^o}2CO_2\)

  2        1         2   (mol)

0,2       0,1      0,2   (mol)

\(4H_2+O_2\underrightarrow{t^o}2H_2O\)

4         1         2     (mol)

0,8       0,2      0,4 (mol)

\(mCO=0,2.28=5,6\left(g\right)\)

\(mH_2=0,8.2=0,16\left(g\right)\)

\(\%mCO=\dfrac{5,6.100}{5,6+0,16}=97,22\%\)

\(\%mH_2=100-97,22=2,78\%\)

nCO2=0,2mol

PTHH: 2CO+O2=>2CO2

         0,2<--0,1<---0,2

=> mO2=0,2.32=6,4g

=> khối lượng Oxi phản ứng với H2 là :

9,6-6,4=3,2g

=> nH2O=3,2:32=0,1mol

PTHH: 2H+O2=>H2O

           0,2<-0,1<-0,2

=> mH2=2.0,2=0,4g

mCO =0,2.28=5,6g

=> m hh=5,6+0,4=6g

a) PTHH: \(2CO+O_2\underrightarrow{t^o}2CO_2\)  (1)

                \(4H_2+O_2\underrightarrow{t^o}2H_2O\)  (2)

b) Ta có: \(\left\{{}\begin{matrix}\Sigma n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\\n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{O_2\left(1\right)}=0,1mol\\n_{O_2\left(2\right)}=0,2mol\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CO}=0,1\cdot28=2,8\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{CO}=\dfrac{2,8}{2,8+0,4}\cdot100\%=87,5\%\\\%m_{H_2}=12,5\%\end{matrix}\right.\)

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

Theo PTHH: \(n_{KMnO_4}=2n_{O_2}=0,6mol\)

\(\Rightarrow m_{KMnO_4}=0,6\cdot158=94,8\left(g\right)\)

nO2 = 9.6/32 = 0.3 (mol) 

nCO2 = 8.8/44 = 0.2 (mol) 

CO + 1/2O2 -to-> CO2 

0.2_____0.1______0.2 

H2 + 1/2O2 -to-> H2O 

0.4__0.3-0.1

%CO = 0.2*28/(0.2*28 + 0.4*2) * 100% = 87.5%

Chúc bạn học tốt !!!

PTHH: \(2CO+O_2\underrightarrow{t^o}2CO_2\)  (1)

            \(2H_2+O_2\underrightarrow{t^o}2H_2O\)  (2)

Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=n_{CO}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\Rightarrow n_{O_2\left(1\right)}=0,1\left(mol\right)\\\Sigma n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{O_2\left(2\right)}=0,2\left(mol\right)\) \(\Rightarrow n_{H_2}=0,4\left(mol\right)\)

\(\Rightarrow\%V_{H_2}=\dfrac{0,4}{0,4+0,2}\cdot100\%\approx66,67\%\)

\(\Rightarrow\%V_{CO}=33,33\%\)

$n_{CO_2} = \dfrac{8,8}{44} = 0,2(mol)$

\(2CO+O_2\xrightarrow[]{t^o}2CO_2\)

0,2       0,1       0,2                 (mol)

$n_{O_2} = \dfrac{9,6}{32} = 0,3(mol)$

\(2H_2+O_2\xrightarrow[]{t^o}2H_2O\)

0,4     0,2         0,2              (mol)

\(\%m_{CO}=\dfrac{0,2.44}{0,2.44+0,4.2}.100\%=91,67\%\\ \%m_{H_2}=100\%-91,67\%=8,33\%\)

\(\%n_{CO}=\dfrac{0,2}{0,2+0,4}.100\%=33,33\%\\ \%n_{H_2}=100\%-33,33\%=66,67\%\)

C₂H₂ + H₂ C₂H₄

C₂H₂ + H₂ C₂H₆

Khi cho hỗn hợp B qua dd nước Br₂ chỉ có C₂H₄ và C₂H₂ phản ứng C₂H₂ +2Br₂ → C₂H₂Br₄

C₂H₄ + Br₂ → C₂H₄Br₂

=> khối lượng bình Br₂ tăng chính bằng khối lượng của C₂H₂ và C₂H₄

m_C2H2 + m_C2H4 = 4,1 (g)    

Hỗn hợp khí D đi ra là CH₄, C₂H₆ và H₂

CH₄ + 2O₂ → t ∘  CO₂ + 2H₂O

C₂H₆ + O₂  → t ∘  2CO₂ + 3H₂O

2H₂ + O₂  → t ∘  2H₂O

Bảo toàn nguyên tố O cho quá trình đốt cháy hh D ta có:

2n_CO2 = 2n_O2 – n_H2O => n_CO2 = ( 0,425. 2– 0,45)/2 = 0,2 (mol)

Bảo toàn khối lượng : m_{hh D} = m_CO2 + m_H2O – m_O2 = 0,2.44 + 0,45.18 – 0,425.32 = 3,3 (g)

Bảo toàn khối lương: m_A = (m_C2H2 + m_C2H4) + m_{hh D} = 4,1 + 3,3 = 7,4 (g)