Question #14b39

The idea here is that you need to use the mole ratio that exists between the reactants and the masses given to you to determine whether or not you're dealing with a limiting reagent.

So, the balanced chemical equation for this reaction looks like this

#"P"_text(4(s]) + color(red)(6)"Cl"_text(2(g]) -> 4"PCl"_text(3(l])#

Notice the #1:color(red)(6)# moel ratio that exists between phosphorus and chlorine. This tells you that the reaction will always consume #color(red)(6)# times more moles of chlorine than of phosphorus.

Use the molar masses of the two reactants to find how many moles of each you have

#45.25color(red)(cancel(color(black)("g"))) * "1 mole P"_4/(123.90color(red)(cancel(color(black)("g")))) = "0.3652 moles P"_4#

and

#130.9color(red)(cancel(color(black)("g"))) * "1 mole Cl"_2/(70.906 color(red)(cancel(color(black)("g")))) = "1.846 moles Cl"_2#

So, do you have enough moles of chlorine gas to allow all the moles of phosphorus to react?

#0.3652color(red)(cancel(color(black)("moles P"_4))) * (color(red)(6)" moles Cl"_2)/(1color(red)(cancel(color(black)("mole P"_4)))) = "2.191 moles Cl"_2#

Notice that you ahve fewer moles of chlorine than you would have needed to make sure that all the moles of phosphorus react. This means that chlorine gas will be a limiting reagent.

More specifically, only

#1.846color(red)(cancel(color(black)("moles Cl"_2))) * "1 mole P"_4/(color(red)(6)color(red)(cancel(color(black)("moles Cl"_2)))) = "0.3077 moles P"_4#

will take part in the rection, the rest will be in excess.

This means that you will be left with

#n_"excess" = 0.3652 - 0.3077 = "0.05750 moles P"_4#

The mass of phosphorus that contains this many moles is

#0.05750color(red)(cancel(color(black)("moles"))) * "123.90 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("7.124 g P"_4)#