∆Hf0 for the formation of rust (Fe2O3) is -826 kJ/mol. How much energy is involved in the formation of 5 grams of rust?

Question

∆Hf0   for the formation of rust (Fe2O3) is -826 kJ/mol. How much energy is involved in the formation of 5 grams of rust?

aaronq · Answer

First, write the a chemical equation for the process.
Convert the mass given (5 grams) to moles.

multiply the moles of $Fe_2O_3$ by $\Delta H_f^{\circ}$

anonymous · Answer

ok is iit 25.9k or 29.5kj? @aaronq

anonymous · Answer

i mean 25.9 kj

aaronq · Answer

it's -25.862 kJ

sweetburger · Answer

@aaronq does converting to moles have anything to do with Hess's Law or am i just wrong by thinking that?

anonymous · Answer

the answer choices are;
25.9kj
25.9j
66kj
66j

@aaronq

aaronq · Answer

if you round up -25.862 kJ, it's -25.9 kJ.

It should also be a negative quantity because it's exothermic.

aaronq · Answer

@sweetburger not really, but i think it's related, it's essentially this formula 

$\Delta H_{rxn}=\Delta H_f^{\circ}*n$

sweetburger · Answer

alright thankyou