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.