# Question 237b1

Nov 4, 2015

$\Delta {H}_{\text{rxn" = -"399.3 kJ}}$

#### Explanation:

I assume that you mistyped the chemical equation for that reaction, since it's pretty clear that $x$ has to be equal to $2$.

So, the balanced chemical equation for your reaction is

$4 {\text{Fe"_text((s]) + 3"O"_text(2(g]) -> 2"Fe"_2"O}}_{\textrm{3 \left(s\right]}}$

Now, the important thing to notice about the standard enthalpy of formation of ferric oxide, ${\text{Fe"_2"O}}_{3}$, is that it's given per mole.

This means that the reaction that forms ferric oxide will give off $\text{826.0 kJ}$ for every mole of ferric oxide formed.

Now, you need to use the molar mass of iron to find how many moles you would get in the $\text{53.99 g}$ sample of iron metal

53.99color(red)(cancel(color(black)("g"))) * " 1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = "0.9668 moles Fe"

Now look at the $4 : 2$ mole ratio that exists between iron metal and ferrioc oxide. This tells you that the reaction will produce $1$ mole of ferric acid for every $2$ moles of iron metal that take part in the reaction.

Using the number of moles of iron metal, the reaction will produce

0.9668color(red)(cancel(color(black)("moles Fe"))) * ("1 mole Fe"_2"O"_3)/(2color(red)(cancel(color(black)("moles Fe")))) = "0.4834 moles Fe"_2"O"_3

So, if the formation of one mole of ferric oxide releases $\text{826.0 kJ}$ of heat, it follows that this reaction will release

0.4834color(red)(cancel(color(black)("moles Fe"))) * "826.0 kJ"/(1color(red)(cancel(color(black)("mole Fe")))) = "399.29 kJ"

Rounded to four sig figs, the heat released by the reaction will be

$q = \text{399.3 kJ}$

This means that the enthalpy change of reaction, $\Delta {H}_{\text{rxn}}$, will be equal to

DeltaH_"rxn" = color(green)(-"399.3 kJ")#

The minus sign is used to designate heat released.