Đáp án $+$ Giải thích các bước giải:

$a)$ $n_{CH_4}=\dfrac{m}{M}=\dfrac{3,2}{16}=0,2$ $(mol)$

$CH_4+2O_2$ $\xrightarrow{t^o}$ $CO_2+2H_2O$
$0,2$    $→$   $0,4$ $→$  $0,2$ $→$ $0,4$

$m_{O_2}=n×M=0,4×32=6,4$ $(g)$

$b)$ $V_{CO_2}=n×22,4=0,2×22,4=4,48$ $(l)$

$c)$ $m_{H_2O}=n×M=0,4×18=7,2$ $(g)$

Cân bằng pt:

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

a)

`n_(CH_4)=(3,2)/16=0,2` (mol)

`n_(O_2)=(0,2.2)/1=0,4` (mol)

`m_(O_2)=0,4.32=12,8` (g)

b)

`n_(CO_2)=(0,2.1)/1=0,2` (mol)

`V_(CO_2)=0,2.22,4=4,48` (l)

c)

`n_(H_2O)=(0,2.2)/1=0,4` (mol)

`m_(H_2O)=0,4.18=7,2` (g)