Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
Putting the values we get :
2 moles of butane releases heat = 5314.8 kJ
1 mole of butane release heat = 
Thus enthalpy of combustion per mole of butane is -2657.4 kJ
<span>d.2HNO3 (aq) + Sr(OH)2 (aq) → 2H2O (l) + Sr(NO3)2(aq)
4H </span>4H
8O 8O
2N 2N
1Sr 1Sr<span>
</span>
Temperature change, colour change, releasing gas, bubbles and change in odor