First, let's consider the reaction :

$$2\mathrm{Al}(s) + \mathrm{Fe}_2\mathrm{O}_3(s) \rightarrow \mathrm{Al}_2\mathrm{O}_3(s) + 2\mathrm{Fe}(s)$$

The heat change for this reaction $$\Delta H^0$$ can be calculated from the heats of formation of the reactants and the products :

$$\Delta H^0 = [\Delta H_f^0(\mathrm{Al}_2\mathrm{O}_3) + 2\Delta H_f^0(\mathrm{Fe})] - [2\Delta H_f^0(\mathrm{Al}) + \Delta H_f^0(\mathrm{Fe}_2\mathrm{O}_3)]$$

Assuming the elements in their standard states have zero enthalpy of formation, i.e.,

$$\Delta H_f^0(\mathrm{Al}) = \Delta H_f^0(\mathrm{Fe}) = 0$$, we can simplify this to :

$$\Delta H^0 = \Delta H_f^0(\mathrm{Al}_2\mathrm{O}_3) - \Delta H_f^0(\mathrm{Fe}_2\mathrm{O}_3)$$

Substitute the given heats of formation into the equation :

$$\Delta H^0 = (-1700\ \mathrm{kJ/mol}) - (-840\ \mathrm{kJ/mol}) = -860\ \mathrm{kJ/mol}$$

This is the heat of reaction for the above reaction. However, we are asked to find the heat evolved per gram of the mixture.

To find this, we need to determine the molar mass of the reactants in the reaction. The molar mass of $$\mathrm{Fe}_2\mathrm{O}_3$$ is $$2 \times 56 + 3 \times 16 = 160\ \mathrm{g/mol}$$ and the molar mass of 2 moles of $$\mathrm{Al}$$ is $$2 \times 27 = 54\ \mathrm{g/mol}$$. The total molar mass of the mixture is $$160 + 54 = 214\ \mathrm{g/mol}$$.

So, the heat evolved per gram of the mixture is $$\frac{-860\ \mathrm{kJ/mol}}{214\ \mathrm{g/mol}} = -4.0187\ \mathrm{kJ/g}$$

Rounded to the nearest integer, this value is approximately -4 kJ/g. The negative sign indicates that the heat is evolved (exothermic reaction).