We must first identify the limiting reactant, and then we calculate the theoretical yield.

We start with the balanced equation.

#"cyclopentadiene + maleic anhydride" → "DA adduct"#

#color(white)(mmmmmmmm)"Cp" + "MA" → "adduct"#

#"MM/g·mol"^(-1): 66.10color(white)(ll) 98.06color(white)(mll) 164.16#

**(a) Identify the limiting reactant**

We calculate the amount of adduct that can form from each reactant.

**Calculate moles of #"Cp"#**

The density of #"Cp"# is 0.786 g/mL.

#"Mass of Cp" = 5.0 color(red)(cancel(color(black)("mL Cp"))) × "0.786 g Cp"/(1 color(red)(cancel(color(black)("mL Cp")))) = "3.93 g Cp"#

#"moles of Cp" = 3.93 color(red)(cancel(color(black)("g Cp"))) × "1 mol Cp"/(66.10 color(red)(cancel(color(black)("g Cp")))) = "0.060 mol Cp"#

Calculate moles of #"adduct"# formed from the #"Cp"#

#0.060color(red)(cancel(color(black)("mol Cp"))) × "1 mol adduct"/(1 color(red)(cancel(color(black)("mol Cp")))) = "0.060 mol Cp"#

**Calculate the moles of #"DA"#**

#"Moles of DA" = 0.015 color(red)(cancel(color(black)("L MA"))) × "4 mol DA"/(1 color(red)(cancel(color(black)("L DA")))) = "0.06 mol DA"#

Calculate moles of #"adduct"# formed from the #"DA"#

#0.06 color(red)(cancel(color(black)("mol DA"))) × "1 mol adduct"/(1 color(red)(cancel(color(black)("mol DA")))) = "0.06 mol adduct"#

This is an equimolar reaction of #"Cp"# and #"DA"#.

There is no limiting reactant.

**(b) Calculate the theoretical yield of #"adduct"#.**

We can use either reactant to calculate the theoretical yield

#"Theoretical yield" = 0.060 color(red)(cancel(color(black)("mol adduct"))) × "164.16 g adduct"/(1 color(red)(cancel(color(black)("mol adduct")))) = "9.8 g adduct"#

The theoretical yield of product is 9.8 g.