Question #42d59

1 Answer
Jan 6, 2016

Answer:

#"520 g"#

Explanation:

Notice that the problem provides you with the thermochemical equation for your reaction.

As you know, the thermochemical equation is simply a balanced chemical equation that includes the enthalpy change of reaction, #DeltaH_"rxn"#

#color(red)(3)"Fe"_text((s]) + 2"O"_text(2(g]) -> "Fe"_3"O"_text(4(s]), " "DeltaH_text(rxn) = -"1120 kJ"#

So, what does this tell you?

When the reaction produces one mole of iron(II, III) oxide, a total of #"1120 KJ"# of heat are being given off #-># you are dealing with an exothermic reaction.

Remember, heat given off is represented by a minus sign attached to the value of #DeltaH_"rxn"#.

Since you know that one mole will give off #"1120 kJ"# of heat, you can say that #"3600 kJ"# of heat will be given off when

#3600 color(red)(cancel(color(black)("kJ"))) * ("1 mole Fe"_3"O"_4)/(1120color(red)(cancel(color(black)("kJ")))) = "3.214 moles Fe"_3"O"_4#

Now all you need to do is figure out how many moles of iron must react in order to produce #3.214# moles of iron(II, III) oxide.

Notice that you have a #color(red)(3):1# mole ratio between iron and iron(II, III) oxide. This tells you that the reaction will always consume three times as many moles of iron than the number of moles of iron(II, III) oxide it produces.

Therefore, you can say that

#3.124 color(red)(cancel(color(black)("moles Fe"_3"O"_4))) * (color(red)(3)" moles Fe")/(1color(red)(cancel(color(black)("mole Fe"_3"O"_4)))) = "9.372 moles Fe"#

Finally, to determine how many grams of iron would contain this many moles, use the element's molar mass

#9.372 color(red)(cancel(color(black)("moles Fe"))) * "55.845 g"/(1color(red)(cancel(color(black)("mole Fe")))) = "523.38 g"#

Rounded to two sig figs, the answer will be

#m_(Fe) = color(green)("520 g")#

So, when #"520 g"# of iron react with excess oxygen, the reaction gives off #"3600 kJ"# of heat.

This is equivalent to saying that when #"520 g"# of iron react with excess oxygen, the enthalpy change of reaction is equal to

#DeltaH_"rxn" = -"3600 kJ"#