`a,`

`PTHH:CH_4+2O_2` $\xrightarrow{t^o}$ `CO_2+2H_2O(1)`

`2H_2+O_2` $\xrightarrow{t^o}$ `2H_2O(2)`

Theo bài ra: `n_{CO_2}=(5,6)/(22,4)=0,25(mol)`

Theo `(1):n_{CH_4}=n_{CO_2}=0,25(mol)`

`=>m_{CH_4}=0,25.16=4(g)`

`->m_{H_2}=5-1=4(g)`

`\%m_{CH_4}=4/(5).100\%=80\%`

`\%m_{H_2}=1/(5).100\%=20\%`

`b,`

`PTHH:H_2+Cl_2` $\xrightarrow{askt}$ `2HCl(3)`

`CH_4+Cl_2` $\xrightarrow{askt}$ `CH_3Cl+HCl(4)`

Theo `(3),(4):n_{Cl_2}=n_{CH_4}+n_{H_2}`

                                `=0,25+1/2=0,75(mol)`

 `=>V_{Cl_2}=0,75.22,4=16,8(l)`

$a)btC: n_{CO_2}=n_{CH_4}=\dfrac{5,6}{22,4}=0,25(mol)$

$\rightarrow n_{H_2}=\dfrac{5-0,25.16}{2}=0,5(mol)$

$\rightarrow \%m_{CH_4}=\dfrac{0,25.16}{5}=80\%;\%m_{H_2}=20\%$
$b) H_2+Cl_2\xrightarrow{as}2HCl$

$CH_4+Cl_2\xrightarrow{as}CH_3Cl+HCl$

theo pt:$n_{Cl_2}=n_{CH_4}+n_{H_2}=0,75(mol)$

$\rightarrow V_{Cl_2}=22,4.0,75=16,8l$