Limiting reactant and theoretical/percent yield

Question

A precipitation reaction occurs when 653 mL of 0.888 M Pb(NO3)2 reacts with319 mL of 0.936 M KI, as shown by the following equation.

Pb(NO3)2(aq) + 2KI(aq) = PbI2(s) +2KNO3(aq)

1. Identify the limiting reactant.
2. Calculate the theoretical yield of PbI2 from the reaction
3. Calculate the percent yield of PbI2 if 50.4g of PbI2 are formed experimentally.

Answer 1

In order to determine the limiting reactant let's first see how much of each reactant is present in the mixture.

For Pb(NO3)2 we have 0.653 L*0.888 moles/L=0.580 moles

For KI we have 0.319L* 0.936 moles/L=0.298 moles.

Now, we look at the stoichiometric relationships from the balanced equation. The equation says we need 2 moles of KI for 1 mole of Pb(NO3)2. Let's assume that ALL Pb(NO3)2 reacted. If that were true than we would need 0.580*2=1.160 mole of KI. But we DO NOT HAVE that much. Therefore, KI is the limiting reactant.

There is another way to answer this question, which is popular among high school teachers.

We calculate how much of the product we would obtain if we used up all available reactants individually. Whichever reactant gives us a smaller amount of product is the limiting reactant.

Let's see what happens here. IF we used up all Pb(NO3)2 we would produce 0.580 moles of PbI2 because balances equation states that 1 mole of Pb(NO3)2 produces 1 mole of PbI2.

If we used up all KI we would produce 0.298/2=0.149 moles of PbI2 because balances equation states than it takes 2 moles of KI to produce 1 mole of PbI2. We see that 0.149 is smaller value than 0.580, hence KI is the limiting reactant.

To calculate theoretical yield, we only focus on the limiting reactant, which in our case is KI. We already calculated that if we used up all KI we would produce 0.149 moles of PbI2 . To convert this into grams we need molar mass of PbI2, which is 461.01 g/mole. We can now calculate the mass of the product: 0.149*461.01g/mol=68.7 g PbI2

Percent yield is then simply a ratio of actual vs theoretical yield (50.4/68.7)*100=73.4%