`n_(O_2)=(3,36)/(22,4)=0,15(mol)`

`=>m_(O_2)=0,15.32=4,8(g)`

 Kết quả cả hai lần cân khối lượng không thay đổi

 `=>m_(O_2)=m_(hh)=4,8(g)` (Hiđro và cacbonic)

Gọi `n_(H_2)=a,n_(CO_2)=b(mol)`

`=>2a+44b=m_(hh)=4,8(1)`

$\overline{M} _{hh}$`=32.0,5=16`$(g/mol)$

`=>(4,8)/(a+b)=16`

`=>16a+16b=4,8(2)`

Từ `(1),(2)=>`$\begin{cases} a=0,2\\b=0,1 \end{cases}$

`=>m_(H_2)=0,2.2=0,4(g)`

`%m_(H_2)=(0,4)/(4,8).100%``=8,33%`

`=>%_(CO_2)=100%-8,33%=91,67%`

`#Ly`

`n_(O_2` `=` `(3,36)/(22,4)=0,15` $(mol)$

`m_(O_2` `=` `0,15.32=4,8` $(g)$

$\overline{M}$ $=$ $0,5.32=16$ $(g/mol)$

Áp dụng sơ đồ đường chéo:
$\rm \text{$n_{H_2}$ $M_{H_2}$ $=$ $2$ } \hspace{5,6cm} \text{$28$} \\\hspace{4cm} \huge{\searrow} \hspace{2,5cm}\huge{\nearrow} \\\hspace{5,5cm}\text{$\overline{M}$ $=$ $16$} \\ \\\hspace{4cm} \huge{\nearrow} \hspace{2,5cm} \huge{\searrow} \\\ \text{$n_{CO_2}$ $M_{CO_2}$ $=$ $44$} \hspace{5,5cm} \text{$14$}$

`=>` $\dfrac{n_{H_2}}{n_{CO_2}}$ $=$ $\dfrac{28}{14}$ $=$ $2$

`=>` `2n_(CO_2` $-$ `n_(H_2` `=` `0`

Mà `44n_(CO_2` $+$ `2n_(H_2` `=` `4,8` $(g)$

$⇒$ $\begin{cases} n_{CO_2}=0,1(mol)\\n_{H_2}=0,2(mol)\end{cases}$

`m_(CO_2` `=` `0,1.44=4,4` $(g)$

`%m_(CO_2` `=` `(4,4)/(4,8).100%≈91,67%`

`%m_(O_2` `=` `100-91,67=8,33%`