Question

Calculate the enthalpy change for the thermite reaction: 2al(s)+fe2o3(s)→2fe(s)+al2o3(s), δh∘rxn=−850 kj when 8.00 mol of al undergoes the reaction with a stoichiometrically equivalent amount of fe2o3. express your answer to three significant figures and include the appropriate units.

Answer 1

Enthalpy change for the balanced equation is -850kj
And from the balanced equation we come to know that for 2 moles of Al, the value is -850kj
for 1 mole of Al = -850/2 = -425 kj
We have to find the enthalpy change for 8 moles of Al = 8 x -425 = 3400kj
Thus, the enthalpy change for 8 mol of Al is -3400kj.

Answer 2

The reaction is

2Al(s) + Fe2o3(s) →2Fe(s) + Al2O3(s)

As given that the enthalpy of reaction is = -850 kJ

This energy is released when two moles of aluminium reacts with one mole of Fe2O3 is -850 kJ

Thus the energy released on reaction of one mole of Al = -850 /2 = -425kJ

The energy released on reaction of eight moles of Al = -425 X 8 kJ

                                                                                           = 3400kJ