$PTHH : C_4H_{10} +\dfrac{13}{2} O_2 \xrightarrow{to}4 CO_2 +5H_2O$ 

$Δ_rH_{298}^0 =4Δ_fH_{298}^0(CO_2) +5Δ_fH_{298}^0(H_2O) -Δ_fH_{298}^0(C_4H_{10})$ 

                         $=4.(-393,51) +5.(-241,82) -(-126,15)=-2656,99(kJ)$

$n_{C_4H_{10}}=\dfrac{2}{58} =\dfrac{1}{29}(mol)$ 

khi đốt `2(g)` `C_4H_{10}` $=\dfrac{1}{29} . (-2656,99)≈-91,62(kJ)$ 

`16)`

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

`->\Delta_r H_{298}^0=4.(-393,51)+5.(-241,82)-(-126,15)=-2656,99(kJ)`

`n_{C_4H_{10}}=2/{58}=1/{29}(mol)`

`->` Lượng nhiệt tỏa ra là `1/{29}.2656,99=91,62(kJ)`