`a)` 

Phương trình hóa học 

$\rm C_3H_8 +5O_2 \xrightarrow{to} 3CO_2 +4H_2O$ 

$\rm C_4{10} +\dfrac{13}{2}O_2 \xrightarrow{to} 4CO_2 +5H_2O$ 

`b)` 

Ta có $\rm C_3H_8:C_4H_{10}= 2:3$ 

Suy ra trong `1` bình gas `12(kg)` có 

$\rm m_{C_3H_8}=4,8(kg)$ 

`->`$\rm n_{C_3H_8}=\dfrac{4,8}{44}=\dfrac{6}{55}(kmol)$ 

$\rm m_{C_4H_{10}}=7,2(kg)$ 

`->`$\rm n_{C_4H_{10}}=\dfrac{7,2}{58}=\dfrac{18}{145}(kmol)$ 

Theo phương trình hóa học 

$\rm n_{O_2}=5n_{C_3H_8}+\dfrac{13}{2}n_{C_4H_{10}}=5.\dfrac{6}{55}.10^3 +\dfrac{13}{2}.\dfrac{18}{145}.10^3=\dfrac{2157}{1595}.10^3(mol)$ 

`->` $\rm V_{không~khí}=\dfrac{\dfrac{2157}{1595}10^3..24,79}{21\%}=159641,8271(l)$