$\text{Ta có số mol H$_{2}$ : n$_{H2}$=3.10$^{23}$:6.10$^{23}$=0,5(mol)}$

$\text{PTHH:}$

$\text{2H$_{2}$+O$_{2}$→2H$_{2}$O}$

$\text{0,5  →0,25  →0,5(mol)}$

$\text{a,m$_{H2O}$=0,5.18=9(g)}$

$\text{b,V$_{H2O}$=0,5.22,4=11,2(lít)}$

$\text{c,V$_{O2 cần}$=0,25.22,4=5,6(lít)}$

$\text{d,Ta có trong không khí có chứa $\frac{1}{5}$thể tích khí O2 }$

$\text{⇒V$_{không khí}$=5,6:$\frac{1}{5}$=28(lít)}$

Câu 5:

`a.` `n_{H_2}``=``\frac{3.10²³}{6.10²³}``=``0,5` `(mol)`

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

                       `0,5`       `0,25`      `0,5`       `mol`

`→``n_{H_2O}``=``n_{H_2}``=``0,5` `(mol)`

`→``m_{H_2O}``=``n``.``M``=``0,5``.``18``=``9` `(g)`

`b.` `V_{H_2O}``=``n``.``22,4``=``0,5``.``22,4``=``11,2` `(l)`

`c.`Theo pt, ta có: `n_{O_2}``=``n_{H_2}``=``\frac{0,5.1}{2}``=``0,25` `(mol)`

`→`$V_{O_2(đktc)}$`=``n``.``22,4``=``0,25``.``22,4``=``5,6` `(l)`

`d.` `V_{kk}``=``V_{O_2}``.``5``=``5,6``.``5``=``28` `(l)`