Question #607c1

The balanced chemical equation that describes your reaction looks like this

#4"Fe"_ ((s)) + 3"O"_ (2(g)) -> 2"Fe"_ 2"O"_ (3(s))#

This tells you that in order for the reaction to take place, you need to provide #3# moles of oxygen gas for every #4# moles of iron that take part in the reaction.

So start by converting the mass of iron and the mass of oxygen gas to moles by using the molar masses of the two reactants.

#56 color(red)(cancel(color(black)("g"))) * "1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = "1.0028 moles Fe"#

#24 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "0.75 moles O"_2#

Now, your goal is to figure out if you have enough moles of one reactant (you can pick either one) to ensure that all the moles of the second reactant will get the chance to take part in the reaction.

For example, in order for #1.0028# moles of iron to react, you need

#1.0028 color(red)(cancel(color(black)("moles Fe"))) * "3 moles O"_2/(4color(red)(cancel(color(black)("moles Fe")))) = "0.7521 moles O"_2#

Since you need more moles of oxygen gas than you have in your sample

#overbrace("0.7521 moles O"_2)^(color(blue)("what you need")) " "> " " overbrace("0.75 moles O"_2)^(color(blue)("what you have"))#

you can say that oxygen gas will be the limiting reagent, i.e. it will be completely consumed before all the moles of iron will get the chance to react.

That said, it's important to realize that you only have #2# significant figures for your values. This means that the number of moles of each reactant should be rounded to two sig figs, in which case you'd have

#"1.0 moles Fe " and " 0.75 moles O"_2#

Thsi time, you can say that #1.0# moles of iron will require

#1.0 color(red)(cancel(color(black)("moles Fe"))) * "3 moles O"_2/(4color(red)(cancel(color(black)("moles Fe")))) = "0.75 moles O"_2#

which is exactly how much oxygen gas you have in your sample. This implies that neither the iron nor the oxygen gas will act as the limiting reagent!

This is actually a great example of how the number of significant figures can influence the answer to a question.

• if you don't take into account the number of sig figs, you can say that oxygen gas is the limiting reagent.
• if you do take into account the number of sig figs, as you should do, you can say that you don't have a limiting reagent here.