Question #237b1

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.