Help me

Đáp án:

Gọi: $\begin{cases}\rm n_{C_7H_{16}}=0,1a\ (mol)\\\rm n_{C_8H_{18}}=0,5a\ (mol)\\\rm n_{C_{9}H_{20}}=0,3a\ (mol)\\\rm n_{C_{10}H_{22}}=0,1a\ (mol)\end{cases}$

Khối lượng xăng:

$\rm m=100a+114×5a+128×3a+142a=2000\ (g)$

$\rm\to a=\dfrac{500}{299}\ (mol)$

Bảo toàn nguyên tố C:

$\rm n_{CO_2}=7a+8×5a+9×3a+10a=140,468\ (mol)$

$\rm\to V_{CO_2}=\dfrac{140,468×0,082×(27,3+273)}{1}=3458,97\ (L)$

$\rm\to Q_t=10a×5337,8×20\%=17852,2\ (kJ)$