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a, 

$C_4H_{10}+\dfrac{13}2O_2\xrightarrow{t^o} 4CO_2+5H_2O$

$\Delta_rH^{0}_{298}=4\Delta_fH^{0}_{298}(CO_2)+5\Delta_fH^{0}_{298}(H_2O)-\Delta_fH^{0}_{298}(C_4H_{10})-\dfrac{13}2\Delta_fH^{0}_{298}(O_2)=-2874,2\ (kJ)$

b,

`n_{C_4H_10}=(8,7)/58=0,15\ (mol)`

`n_{O_2}=32/32=1\ (mol)`

Vì `n_{C_4H_10}/1=0,15<n_{O_2}/(6,5)=0,154`

`\to C_4H_10` pứ hết 

1 mol `C_4H_10` cháy tỏa ra 2874,2 kJ

`\to` 0,15 mol `C_4H_10` cháy tỏa ra `2874,2.0,15=431,13` kJ