Question

Question #237b1

Answer 1

`DeltaH_"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"Fe"_text((s]) + 3"O"_text(2(g]) -> 2"Fe"_2"O"_text(3(s])`

Now, the important thing to notice about the standard enthalpy of formation of ferric oxide, `"Fe"_2"O"_3`, is that it's given per mole.

This means that the reaction that forms ferric oxide will give off `"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 `"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 `"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 = "399.3 kJ"`

This means that the enthalpy change of reaction, `DeltaH_"rxn"`, will be equal to

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

The minus sign is used to designate heat released.