Question #8b1ae

You know that the balanced chemical equation that describes this reaction looks like this

#6"Li"_ ((s)) + "N"_ (2(g)) -> 2"Li"_ 3"N"_ ((s))#

Notice that the reaction requires #6# moles of lithium for every #1# mole of nitrogen gas that takes part in the reaction.

Start by using the molar mass of lithium to convert the mass to moles.

#14.6 color(red)(cancel(color(black)("g"))) * "1 mole Li"/(6.941color(red)(cancel(color(black)("g")))) = "2.1034 moles Li"#

Do the same for nitrogen gas.

#24.2 color(red)(cancel(color(black)("g"))) * "1 mole N"_2/(28.0134color(red)(cancel(color(black)("g")))) = "0.86387 moles N"_2#

Next, pick one of the two reactants and check to see if you have enough moles in your sample to allow for all the moles of the second reactant to take part in the reaction.

Let's pick nitrogen gas this time. Use the #6:1# mole ratio that exists between lithium and nitrogen gas to see how many moles of lithium would be required to consume all the moles of nitrogen gas.

#0.86387 color(red)(cancel(color(black)("moles N"_2))) * "6 moles Li"/(1color(red)(cancel(color(black)("mole N"_2)))) = "5.1832 moles Li"#

As you can see, your sample does not contain that many moles of lithium

#overbrace("2.1034 moles Li")^(color(blue)("what you have")) " " < " " overbrace("5.1832 moles Li")^(color(blue)("what you need"))#

which means that lithium will act as the limiting reagent, i.e. it will be completely consumed before all the moles of nitrogen gas will get the chance to react.

In other words, nitrogen gas is in excess here because you have more nitrogen gas than you'd need to consume all the moles of lithium present in your sample.

#2.1034 color(red)(cancel(color(black)("moles Li"))) * "1 mole N"_2/(6color(red)(cancel(color(black)("moles Li")))) = "0.33557 moles N"_2#

This time, you have more nitrogen gas than you need because

#overbrace("0.86387 moles N"_2)^(color(blue)("what you have")) " " > " " overbrace("0.33557 moles N"_2)^(color(blue)("what you need"))#

so nitrogen gas is in excess, which is, of course, equivalent to saying that lithium is the limiting reagent.