`n_{H_2}=(4,48)/(22,4)=0,2(mol)`

a,PTHH:

`2H_2+O_2`$\xrightarrow[]{t^o}$`2H_2O`

`0,2->0,1->0,2(mol)`

Từ PTHH :

`n_{O_2}=0,1(mol)`

`n_{H_2O}=0,2(mol)`

`->m_{H_2O}=0,2.18=3,6(g)`

b,

Ta có :

`V_{O_2}=0,1.22,4=2,24(lít)`

`V_{O_2}=20\%V_{kk}`

`->V_{kk}=V_{O_2}:20\%=2,24:20\%=11,2(lít)`

Đáp án + Giải thích các bước giải:

$a\bigg)$

$n_{H_2}=\frac{4,48}{22,4}=0,2(mol)$

$PTHH:2H_2+O_2\xrightarrow{t^o}2H_2O$

Theo PT: $n_{H_2O}=n_{H_2}=0,2(mol)$

$\to m_{H_2O}=0,2.18=3,6(g)$

$b\bigg)$

Theo PT: $n_{O_2}=\frac{1}{2}n_{H_2}=0,1(mol)$

$\to V_{O_2(đktc)}=0,1.22,4=2,24(l)$

$\to V_{\rm không\,khí(đktc)}=\frac{2,24}{20\%}=11,2(l)$