1) c)solida
2) C4H10+13/2O2=4CO2+5H2O ( /-impartire)
dH=(4dH CO2+ 5dH H2O)-(dH C4H10+ 13/2dH O2)
dH=4(-393,5)+5(-285,8)-(-126+ 13/2(0) )
dH= -1574-1429+126
dH= -2877 kj/mol
1 m3(metru cub)= 1000 l
1 mol C4H10=22,4 L ........-2877 kj
0,224 m3 = 224 L...........x
x=-28770 kj => Q= +28770 kj