Question

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:
$H_{f\left({Al}_2{O}_3\right)}=-399kcal{mol}^{-1};H_{f\left({Fe}_2{O}_3\right)}=-199cal{mol}^{-1}$
Density of ${Fe}_2{O}_3=5.2g/{cm}^3$; Density of $Al=2.7g/{cm}^3$.

Answer 1

The required equation is:
$2Al+{Fe}_2{O}_3⟶{Al}_2{O}_3+2Fe;\Delta H=?$
$\Delta H^{}=\Delta H_{f\left(products\right)}-\Delta H_{f\left(reactants\right)}$
$=[\Delta H_{\left({Al}_2{O}_3\right)}+2\Delta H_{\left(Fe\right)}]-[2\Delta H_{\left(Al\right)}+\Delta H_{\left({Fe}_2{O}_3\right)}]$
$=[-399+2\times 0]-[2\times 0+(-199\left)\right]=-200kcal$
At. mass of aluminium $=27$, Mol. mass of ${Fe}_2{O}_3=160$
Volume of reactants$=\frac{160}{5.2}+\frac{2\times 27}{2.7}=50.77{cm}^3$
Fuel value per ${cm}^3=\frac{200}{50.77}=3.92kcal$