TT
$m_{CH_{4}}=6,4g$
-----------------
a) $m_{CO_{2}}=?$
$m_{H_{2}O}=?$
b) $V_{O_{2}}=?$
Giải
a) Ta có : $n_{CH_{4}}=\frac{m}{M}=\frac{6,4}{16}=0,4(mol)$
PTHH: $CH_{4}+2O_{2}→CO_{2}+2H_{2}O$
`1` `2` `1` `2`
`0,4` `0,8` `0,4` `0,8` (mol)
Mà $n_{CO_{2}}=0,4$
$⇒m_{CO_{2}}=n.M=0,4.44=17,6(g)$
$m_{H_{2}O}=n.M=0,8.18=14,4(g)$
b) Ta có : $n_{O_{2}}=0,8$
$⇒V_{O_{2}}=n.22,4=0,8.22,4=17,92(l)$