( this iron (lll) oxide is the common rust )

for the products

Al2O3 I solid and iron is molten i-e liquid because it was mentioned that the reaction is highly exothermic ……

b) mass of iron (lll) oxide = 10g
realtive formula mass of iron (lll) oxide = 56×2 + 16×3
= 160
so,
moles of iron (lll)oxide = 10/160
= 0.0625 moles

keeping in view the stochiometric ratios of the reactants and the products we get that to react with one mole of Fe2O3 we need two moles of Al

so moles of Al used = 2x 0.0625
=.125 moles

now mass of Al = ?
= moles x molar mass
= .125 x 13
= 1.625 g

c) energy released in bond formation is more than the energy used in bond breaking

d) iron (lll) is gaining electrons and changing from Fe +3 to Fe 0….
as gain of electrons reduction so we term it as iron is reduced