An intimate mixture of ferric oxide and aluminium is used as solid fuel in rockets. Calculate the fuel value per `cm^(3)` of the mixture. Heats of formation and densities are as follows:
`H_(f(Al_(2)O_(3)))=-399kcal" "mol^(-1),H_(f(Fe_(2)O_(3)))=-199kcal" "mol^(-1)` ltbr. Density of `Fe_(2)O_(3)=5.2g//cm^(3)," Density of "Al=2.7g//cm^(3)`

Explanation

The required equation is:
`2Al+Fe_(2)O_(3)toAl_(2)O_(3)+2Fe,DeltaH=?`
`DeltaH=DeltaH_(f("products"))-DeltaH_(f("reactants"))`
`=[DeltaH_(f(Al_(2)O_(3)))+2DeltaH_(f(Fe))]-[2DeltaH_(f(Al))+DeltaH_(f(Fe_(2)O_(3)))]`
`=(-399+2xx0)-[2xx0+(-199)]`
`=-399+199=-200kcal`
At. mass of aluminium `=27,` mol. mass of `Fe_(2)O_(3)=160`
Volume of reactants `=(160)/(5.2)+(2xx27)/(2.7)=50.77cm^(3)`
Fuel value per `cm^(3)=(200)/(50.77)=3.92kcal`