Đáp án:
$n_{H_{2}}$ = $\frac{V}{22,4}$ = $\frac{2,8}{22,4}$ = 0,125 ( mol )
PTHH: 2$H_{2}$ + $O_{2}$ → 2$H_{2}$O (1)
Theo ( 1) ta có:
$n_{O_{2}}$ = $\frac{1}{2}$ $n_{H_{2}}$ = $\frac{1}{2}$. 0.125 = 0,0625 ( mol )
$m_{O_{2}}$ = n. M = 0,0625 . 32 = 2 ( g )
$V_{O_{2}}$ = n . 22.4 = 0,0625 . 22,4 = 1,4 ( l )
Theo ( 1) ta có:
$n_{H_{2}O}$ = $n_{H_{2}}$ = 0,125 ( mol )
$m_{H_{2}O}$ = n . M = 0.125 . 18 = 2,25 ( g )