a)

$CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$

b) Theo PTHH : $V_{O_2} = 2V_{CH_4} = 11,2(lít)$

$n_{CH_4} = \dfrac{5,6}{22,4} = 0,25(mol)$

Theo PTHH : $n_{H_2O} = 2n_{CH_4} = 0,5(mol)$
$m_{H_2O} = 0,5.18 = 9(gam)$

\(a,CH_4+2O_2\rightarrow CO_2+2H_2O\)

\(1:2:1:2\left(mol\right)\)

\(0,25:0,5:0,25:0,5\left(mol\right)\)

\(n_{CH_4}=\dfrac{V}{22,4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(b,V_{O_2}=n.22,4=0,5.22,4=11,2\left(l\right)\)

\(m_{H_2O}=n.M=0,5.18=9\left(g\right)\)

Gọi nCH4 = x (mol), nH2 = y (mol)   (x,y>0)

ta có :
nhỗn hợp = 11,2/22,4 = 0,5 mol

--> x + y = 0,5 (1)

pthh:

CH4 +2O2 ----> CO2 + 2H2O

  x   -->               x     ---> 2x

2H2 + O2 ---> 2H2O

y   --->                y

Lại có: nH2O = 16,2/18 = 0,9 (mol)

----> 2x + y = 0,9 (2)

Từ (1) và (2) ---> x = 0,4            y = 0,1   (mol)

VCH4 = 0,4 . 22,4 = 8,96 l

VH2 = 0,1 . 22,4 = 2,24 l

VCO2 = x.22,4 = 0,4 . 22,4 = 8,96 l

Xong rồi đok

Ta có: \(n_{C_2H_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

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

a, \(n_{O_2}=3n_{C_2H_4}=1,5\left(mol\right)\Rightarrow V_{O_2}=1,5.22,4=33,6\left(l\right)\)

\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=168\left(l\right)\)

b, \(n_{CO_2}=2n_{C_2H_4}=1\left(mol\right)\Rightarrow m_{CO_2}=1.44=44\left(g\right)\)

\(n_{C_2H_6}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: \(2C_2H_6+7O_2\xrightarrow[]{t^o}4CO_2+6H_2O\)

             0,25--->0,875

`=> V_{O_2} = (0,875.22,4)/(20%) = 98 (l)`

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

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

Gọi \(\left\{{}\begin{matrix}n_{CH_4}=x\\n_{H_2}=y\end{matrix}\right.\)

\(n_{O_2}=\dfrac{7,437}{24,79}=0,3mol\)

\(n_{H_2O}=\dfrac{7,2}{18}=0,4mol\)

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

  x        2x                  x            2x      ( mol )

\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)

 y       1/2 y               y        ( mol )

Ta có:

\(\left\{{}\begin{matrix}2x+\dfrac{1}{2}y=0,3\\2x+y=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

\(m=m_{CH_4}+m_{H_2}=\left(0,1.16\right)+\left(0,2.2\right)=2g\)

\(V_{CO_2}=0,1.24,79=2,479l\)

\(M_{hỗn\ hợp} = 4,5.2 = 9\\ Gọi : n_{CH_4} = a(mol) ; n_{H_2} = b(mol)\\ \Rightarrow 16a + 2b =9(a + b)\ (1) n_{O_2} = \dfrac{56}{5.22,4} = 0,5(mol)\\ CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ n_{O_2} = 2a + 0,5b = 0,5(2)\\ (1)(2) \Rightarrow a = 0,2 ; b = 0,2\\ \Rightarrow V = (0,2 + 0,2).22,4 = 8,96(lít)\)

Đề không hợp lí em ơi, CO₂ và H₂O đều không cháy được nhé.

\(n_X=\dfrac{0,896}{22,4}=0,08\left(mol\right)\)

\(M_X=21.2=42\left(g\text{/}mol\right)\\ \rightarrow m_X=0,08.42=3,36\left(g\right)\)

PTHH: 

\(C_3H_4+4O_2\xrightarrow[]{t^o}3CO_2+H_2O\\ 2C_3H_6+9O_2\xrightarrow[]{t^o}6CO_2+6H_2O\\ C_3H_8+5O_2\xrightarrow[]{t^o}3CO_2+4H_2O\)

Theo PTHH: \(n_C=n_{CO_2}=3n_X=3.0,08=0,24\left(mol\right)\)

\(\rightarrow V_{CO_2}=0,24.22,4=5,376\left(l\right)\)

BTNT: 

\(m_H=m_X=m_C=3,36-0,24.12=0,48\left(g\right)\\ \rightarrow n_H=\dfrac{0,48}{1}=0,48\left(mol\right)\)

Theo PTHH: \(n_{H_2O}=\dfrac{1}{2}n_H=\dfrac{1}{2}.0,48=0,24\left(mol\right)\)

\(\rightarrow m_{H_2O}=0,24.18=3,42\left(g\right)\)

nX=0,89622,4=0,08(mol)nX=0,89622,4=0,08(mol)

${}^{}=21.2=42\left(g\text{/}mol\right)\to {}^{}=0,08.42=3,36\left(g\right)$

PTHH: 

${}^{}{}^{}+4{}^{}{}^{}3C{}^{}+{}^{}O2{}^{}{}^{}+9{}^{}{}^{}6C{}^{}+6{}^{}O{}^{}$

Ta có: \(n_{CH_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

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

Theo PT: \(n_{O_2}=2n_{CH_4}=0,4\left(mol\right)\)

\(\Rightarrow m_{O_2}=0,4.32=12,8\left(g\right)\)

\(PTHH:2CH_4+O_2\rightarrow2CO+4H_2\)

               \(2:1:2:4\left(mol\right)\)

               \(0,2:0,1:0,2:0,4\left(mol\right)\)

\(n_{CH_4}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(m_{O_2}=n.M=0,1.\left(16.2\right)=3,2\left(g\right)\)