The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to product. Consider the reaction: P4(s) + 6 Cl2(g) -*4 PCI3(1) If 11.18 g P4 is mixed with 18.56 g Cl2, calculate the theoretical yield (g) of PCl3 produced by the reaction. Solution Number of moles of P4 = 11.18 g / 123.90 g/mol = 0.0902 mol Number of moles of Cl2 = 18.56 g / 70.9060 g/mol = 0.262 mole From the balanced equation we can say that 1 mole of P4 requires 6 mole of Cl2 so 0.0902 mole of P4 will require = 0.0902 mole of P4 *(6 mole of Cl2 / 1 mole of P4) = 0.541 mole of Cl2 But we have only 0.262 mole of Cl2 which is in short so Cl2 is the limiting reactant 6 mole of Cl2 produces 4 mole of PCl3 so 0.262 mole of Cl2 will produce = 0.262 mole of Cl2 *(4 mole of PCl3 / 6 mole of Cl2) = 0.175 mole of PCl3 mass of 1 mol of PCl3 = 137.33 g so the mass of 0.175 mole of PCl3 = 24.0 g Therefore, theoretical yield of PCl3 = 24.0 g .

1. The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is
completely converted to product. Consider the reaction: P4(s) + 6 Cl2(g) -*4 PCI3(1) If 11.18 g
P4 is mixed with 18.56 g Cl2, calculate the theoretical yield (g) of PCl3 produced by the
reaction.
Solution
Number of moles of P4 = 11.18 g / 123.90 g/mol = 0.0902 mol
Number of moles of Cl2 = 18.56 g / 70.9060 g/mol = 0.262 mole
From the balanced equation we can say that
1 mole of P4 requires 6 mole of Cl2 so
0.0902 mole of P4 will require
= 0.0902 mole of P4 *(6 mole of Cl2 / 1 mole of P4)
= 0.541 mole of Cl2
But we have only 0.262 mole of Cl2 which is in short so Cl2 is the limiting reactant
6 mole of Cl2 produces 4 mole of PCl3 so
0.262 mole of Cl2 will produce
= 0.262 mole of Cl2 *(4 mole of PCl3 / 6 mole of Cl2)
= 0.175 mole of PCl3
mass of 1 mol of PCl3 = 137.33 g
so the mass of 0.175 mole of PCl3 = 24.0 g
Therefore, theoretical yield of PCl3 = 24.0 g