TT

$m_{CH_{4}}=6,4g$

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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 : $n_{O_{2}}=0,8$

$⇒V_{O_{2}}=n.22,4=0,8.22,4=17,92(l)$

$n$ $CH_{4}$ = $\frac{m}{M}$ = $\frac{6,4}{16}$ = $0,4$ ( $mol$ )

$CH_{4}$ + $2O_{2}$ → $CO_{2}$ ↑ + 2$H_{2}$ $O$

$Tỷ$ $Lệ$ $Mol$ $các$ $chất$ : $0,4$ ; $0,8$

$a$ ) $m$ $CO_{2}$ = $n. M$ = $0,4.44$ = $17,6$ ( $g$ )

m$H_{2}$O = $n.M$ = $0,8.18$ = $14,4$ ( $g$ )

$b$ ) $VO_{2}$ = $n.22,4$ = $0,8.22,4$ = $17,92$ ( $l$ )