Đặt $n_{C_4H_{10}}=x(mol);n_{C_3H_8}=y(mol)$

Đổi $15kg=15000g$

$\Rightarrow 58x+44y=15000$

$M_{hh}=24,8.M_{H_2}=24,8.2=49,6(g/mol)$

$\Rightarrow n_{hh}=\dfrac{15000}{49,6}\approx 302,42(mol)$

$\Rightarrow x+y=302,42(2)$

Từ $(1)(2)\Rightarrow x=\dfrac{3750}{31};y=\dfrac{5625}{31}$

$PTHH:2C_4H_{10}+13O_2\xrightarrow{t^o}8CO_2\uparrow+10H_2O$

$C_3H_8+5O_2\xrightarrow{t^o}3CO_2\uparrow+4H_2O$

Theo PT: $\Sigma n_{O_2}=6,5x+5y=\dfrac{52500}{31}(mol)$

$\Rightarrow n_{O_2(tt)}=\dfrac{52500}{31}.90\%=\dfrac{47250}{31}(mol)$

$\Rightarrow V_{O_2}=\dfrac{47250}{31}.22,4\approx 34141(lít)$

$\Rightarrow V_{kk}=\dfrac{34141}{20\%}=170705(lít)$