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.
