\(PTHH:CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ Đặt:a=n_{CH_4}\left(mol\right);b=n_{C_2H_4}\left(mol\right)\left(a,b>0\right)\\ Có:\left\{{}\begin{matrix}22,4a+22,4b=8,96\\22,4a+44,8b=11,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,3\\b=0,1\end{matrix}\right.\\ \%V_{CH_4}=\dfrac{a}{a+b}.100\%=\dfrac{0,3}{0,3+0,1}.100\%=75\%;\%V_{C_2H_4}=100\%-75\%=25\%\)

a.\(n_{Br_2}=\dfrac{48}{160}=0,3mol\)

\(n_{hh}=\dfrac{7,84}{22,4}=0,35mol\)

\(C_2B_2+2Br_2\rightarrow C_2H_2Br_4\)

 0,15       0,3                      ( mol )

\(\%V_{C_2H_2}=\dfrac{0,15}{0,35},100=42,85\%\)

\(\%V_{CH_4}=100\%-42,85\%=57,15\%\)

b.\(n_{CH_4}=0,35-0,15=0,2mol\)

\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)

 0,2                                          0,4    ( mol )

\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\)

  0,15                                          0,15      ( mol )

\(m_{H_2O}=\left(0,4+0,15\right).18=9,9g\)

C2H2 :)?

a, nBr2 = 8/160 = 0,05 (mol)

PTHH: C2H4 + Br2 -> C2H4Br2

Mol: 0,05 <--- 0,05 <--- 0,05

Vhh khí = 2,8/22,4 = 0,125 (mol)

%VC2H4 = 0,05/0,125 = 40%

%CH4 = 100% - 40% = 60%

b, nCH4 = 0,125 - 0,05 = 0,075 (mol)

PTHH: C2H4 + 3O2 -> (t°) 2CO2 + 2H2O

Mol: 0,05 ---> 0,15

CH4 + 2O2 -> (t°) CO2 + 2H2O

Mol: 0,075 ---> 0,15

Vkk = (0,15 + 0,15) . 5 . 22,4 = 33,6 (l)

a, Ta có \(n_{CH_4}+n_{C_2H_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\left(1\right)\)

PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

Theo PT: \(n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,05\left(mol\right)\\n_{C_2H_4}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}V_{CH_4}=0,05.22,4=1,12\left(l\right)\\V_{C_2H_4}=0,1.22,4=2,24\left(l\right)\end{matrix}\right.\)

b, \(m_{CH_4}=0,05.16=0,8\left(g\right)\)

c, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

Theo PT: \(n_{Br_2}=n_{C_2H_4}=0,1\left(mol\right)\Rightarrow V_{ddBr_2}=\dfrac{0,1}{1}=0,1\left(l\right)\)

\(m_{C_2H_4}=0,1.28=2,8\left(g\right)\)

Câu 1: Hoàn thành các PTHH sau:

a. CH₄ + 2O₂ to→CO2+2H2O

b. CH₄ + Cl₂ ás→CH3Cl+HCl

c. C₂H₄ +3 O₂ to→2CO2+2H2O

d. C₂H₄ + Br₂ →C2H4Br2

e. nC₂H₄  to→-(-CH2-CH2-)-_n

f. C₂H₂ + Br₂ →C2H2Br2

g. C₂H₂ + \(\dfrac{5}{2}\)O₂ to→2CO2+H2O

h.C₂H₂ + 2Br₂ →C2C2Br4

Theo mình bạn nên tách ra thành nhiều câu và sẽ thuận tiện hơn.

\(Đặt:n_{CH_4}=a\left(mol\right);n_{C_2H_4}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ \Rightarrow\left\{{}\begin{matrix}a+b=0,25\\2a+3b=0,625\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,125\\b=0,125\end{matrix}\right.\\ a,m_{hh}=m_{CH_4}+m_{C_2H_4}=16.0,125+28.0,125=5,5\left(g\right)\\ b,V_{CO_2\left(đktc\right)}=22,4.\left(a+2b\right)=8,4\left(l\right)\)

Gọi số mol CH₄, C₂H₄ là a, b (mol)

=> a + b = \(\dfrac{2,8}{22,4}=0,125\left(mol\right)\) (1)

\(n_{O_2}=\dfrac{7,28}{22,4}=0,325\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

              a----->2a

             C₂H₄ + 3O₂ --t^o--> 2CO₂ + 2H₂O

                b----->3b

=> 2a + 3b = 0,325 (2)

(1)(2) => a = 0,05 (mol); b = 0,075 (mol)

=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,05}{0,125}.100\%=40\%\\\%V_{C_2H_4}=\dfrac{0,075}{0,125}.100\%=60\%\end{matrix}\right.\)

nhh khí = 2,8/22,4 = 0,125 (mol)

Gọi nCH4 = a (mol); nC2H4 = b (mol)

a + b = 0,125 (1)

nO2 = 7,28/22,4 = 0,325 (mol)

PTHH:

CH4 + 2O2 -> (t°) CO2 + 2H2O

a ---> 2a ---> a ---> 2a

C2H4 + 3O2 -> (t°) 2CO2 + 2H2O

b ---> 3b ---> 2b ---> 2b

2a + 3b = 0,325 (2)

(1)(2) => a = 0,05 (mol); b = 0,075 (mol)

%VCH4 = 0,05/0,125 = 40%

%VC2H4 = 100% - 40% = 60%