How do you determine the limiting reagent and what is the theoretical yield of the following problem?

The stoichiometric coefficients that appear in the balanced chemical equation tell you the mole ratios in which the chemical species that take part in the reaction find themselves.

In your case, you have

#"PCl"_ (5(s)) + color(red)(4)"H"_ 2"O"_ ((l)) -> "H"_ 3 "PO"_ (4(aq)) + 5 "HCl"_ ((aq))#

This tells you that the reaction consumes #color(red)(5)# moles of water for every mole of phosphorus pentachloride that takes part in the reaction.

Now, your goal when figuring out if you're dealing with a limiting reagent is to pick a reactant and see if you have enough of the second reactant to allow for all the moles of the first reactant to take part in the reaction.

Let's pick phosphorus pentachloride first. In order for #0.360# moles of phosphorus pentachloride to react, you need to have

#0.360 color(red)(cancel(color(black)("moles PCl"_5))) * (color(red)(4)color(white)(a)"H"_2"O")/(1color(red)(cancel(color(black)("mole PCl"_5)))) = "1.44 moles H"_2"O"#

You supply the reaction with #2.88# moles of water, more than enough to allow for all the moles of phosphorus pentachloride to react.

This means that phosphorus pentachloride will act as a limiting reagent, i.e. it will be completely consumed before all the moles of water get the chance to react.

In other words, water will be in excess.

You get the same result by picking water. In this case, #2.88# moles of water would require

#2.88color(red)(cancel(color(black)("moles H"_2"O"))) * ("1 mole PCl"_5)/(color(red)(4)color(red)(cancel(color(black)("moles H"_2"O")))) = "0.720 moles PCl"_5#

Since you only have #0.360# moles of phosphorus pentachloride available, this will be your limiting reagent.

Now, the theoretical yield corresponds to the amount of product produced if all the moles of reactants that actually react end up producing moles of product.

Notice that the reaction produces #1# mole of phosphoric acid, #"H"_2"PO"_4#, and #5# moles of hydrochloric acid, for #1# mole of phosphorus pentachloride and #color(red)(5)# moles of water.

Since all the moles of phosphorus pentachloride react, the reaction will produce

#0.360 color(red)(cancel(color(black)("moles PCl"_5))) * ("1 mole H"_3"PO"_4)/(1color(red)(cancel(color(black)("mole PCl"_5)))) = "0.360 moles H"_3"PO"_4#

and

#0.360 color(red)(cancel(color(black)("moles PCl"_5))) * "5 moles HCl"/(1color(red)(cancel(color(black)("mole PCl"_5)))) = "1.80 moles HCl"#

This will be the reaction's theoretical yield.