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

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

Ta có: \(n_{CH_4}+n_{C_2H_4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(1\right)\)

Theo PT: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\left(2\right)\)

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

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,2.22,4}{6,72}.100\%\approx66,67\%\\\%V_{C_2H_4}\approx33,33\%\end{matrix}\right.\)

b, Theo PT: \(n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=0,7\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,7.22,4=15,68\left(l\right)\)

\(M_A=1,8125.32=58\left(\dfrac{g}{mol}\right)\\ \rightarrow\left\{{}\begin{matrix}m_C=58.82,76\%=48\left(g\right)\\m_H=58-48=10\left(g\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}n_C=\dfrac{48}{12}=4\left(mol\right)\\n_H=\dfrac{10}{1}=10\left(mol\right)\end{matrix}\right.\\ CTHH:C_4H_{10}\)

\(n_{C_4H_{10}}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: 2C₄H₁₀ + 13O₂ --to--> 8CO₂ + 10H₂O

             0,2                               0,8

=> V_CO2 = 0,8.22,4 = 17,92 (l)

a, \(CH_4+2O_2\underrightarrow{^{t^o}}CO_2+2H_2O\)

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

b, Gọi: \(\left\{{}\begin{matrix}n_{CH_4}=x\left(mol\right)\\n_{C_2H_4}=y\left(mol\right)\end{matrix}\right.\) \(\Rightarrow x+y=\dfrac{4,48}{22,4}=0,2\left(mol\right)\left(1\right)\)

Theo PT: \(n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=2x+3y=\dfrac{15,68}{22,4}=0,7\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=-0,1\\y=0,3\end{matrix}\right.\)

Đến đây thì ra số mol âm, bạn xem lại đề nhé.

\(n_{C_2H_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(n_{O_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)

      \(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t^0}2CO_2+H_2O\)

\(Bđ:0.3.......0.5\)

\(Pư:0.2........0.5.........0.4.........0.2\)

\(Kt:0.1..........0..........0.4...........0.2\)

\(V_{CO_2}=0.4\cdot22.4=8.96\left(l\right)\)

\(V_{C_2H_2\left(dư\right)}=0.1\cdot22.4=2.24\left(l\right)\)

Gọi \(\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\end{matrix}\right.\left(đk:0< a,b< 0,3\right)\)

\(n_{hh}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{O_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)

PTHH:

CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

a------>a

C₂H₄ + 3O₂ --t^o--> 2CO₂ + 2H₂O

b------->3b

=> Hệ pt \(\left\{{}\begin{matrix}a+b=0,3\\a+3b=0,7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\left(TM\right)\)

\(\rightarrow\left\{{}\begin{matrix}V_{CH_4}=0,1.22,4=2,24\left(l\right)\\V_{C_2H_4}=0,2.22,4=4,48\left(l\right)\end{matrix}\right.\)

a)

$n_{Br_2} = \dfrac{160.15\%}{160} = 0,15(mol)$
$C_2H_4 + Br_2 \to C_2H_4Br_2$

Ta thấy : $n_{C_2H_4} = 0,2 > n_{Br_2} = 0,15$ nên $C_2H_4$ dư

$n_{C_2H_4Br_2} = n_{Br_2} = 0,15(mol) \Rightarrow m_{C_2H_4Br_2} = 0,15.188 = 28,2(gam)$

b) $n_{C_2H_4\ dư} = 0,2 - 0,15 = 0,05(mol) \Rightarrow V_{C_2H_4} = 0,05.22,4 = 1,12(lít)$

$C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O$

Theo PTHH : 

$V_{CO_2} =2 V_{C_2H_4} = 2,24(lít)$
$V_{O_2} = 3V_{C_2H_4} = 3,36(lít) \Rightarrow V_{kk} = 5V_{O_2} = 16,8(lít)$