Fe\(_2\)O\(_{3(s)}\) + 2Al\(_{(s)}\) → Al\(_2\)O\(_3\) + 2Fe\(_{(s)}\). If the heats of formation of Al\(_2\)O\(_3\) and Fe\(_2\)O\(_3\) are - 1670 kJmol\(^{-1}\) and - 822 kJmol\(^{-1}\) respectively, the enthalpy change in kJ for the reaction is ?

MhizDee430

10 months ago

I'm shocked 😳
How many persons are actually doing the calculation???
-1670+882 =-788
So where is this -848 coming from??

Nuelzyyy

11 years ago

Delta H(energy changes) =heat of products - heat content of reactants .....



Going with dat....Energy cahange =(-1670) - (-822) = -848

Alkali2

2 months ago

how do we no that it is change in heat

ochonmaifeoma

4 months ago

where is the _848 coming from

ochonmaifeoma

4 months ago

I understand
you did 1670_822=848

Roycee

7 years ago

How do i know what figure is for the reactant and that which is for the product

saladin

6 years ago

the answer is wrong, change in heat of (products - reactants)