2,

PTHH: $\text{4P + $5O_{2}$ $\xrightarrow{t^o}$ $2P_{2}$$O_{5}$ }$

$\text{$n_{P}$ = $\dfrac{m}{M}$ = $\dfrac{62}{31}$ = 2 mol}$

Theo PTHH ta thấy: $\text{$n_{O}$ = $\dfrac{5}{4}$. $n_{P}$ = 2,5 mol}$

$\rightarrow$ $\text{$V_{O_2(đktc)}$ = n. 22,4 = 2,5. 22,4 = 56 lít}$

$\rightarrow$ $\text{$V_{kk}$ = 56 : 20% = 280 lít}$

3,

PTHH: $\text{C + $O_{2}$ $\xrightarrow{t^o}$ $CO_{2}$ }$

$\text{$n_{C}$ = $\dfrac{m}{M}$ = $\dfrac{3,6}{12}$ = 0,3 mol }$

Theo PTHH ta thấy: $\text{$n_{O_2(p.ư)}$ = $n_{C}$ = 0,3 mol}$

$\rightarrow$ $\text{$V_{O_2 (đktc)}$ = n. 22,4 = 0,3. 22,4 = 6,72 lít}$

$\text{#Kudo}$

VD2:

$PTHH:4P+5O_2\xrightarrow{t^o}2P_2O_5$

$n_P=\dfrac{62}{31}=2(mol)$

Theo PT: $n_{O_2}=2,5(mol)$

$\Rightarrow V_{O_2}=2,5.22,4=56(lít)$

$\Rightarrow V_{kk}=\dfrac{56}{20\%}=280(lít)$

VD3:

$PTHH:C+O_2\xrightarrow{t^o}CO_2$

Theo PT: $n_{O_2}=n_C=\dfrac{3,6}{12}=0,3(mol)$

$\Rightarrow V_{O_2}=0,3.22,4=6,72(lít)$