Solid lead(II) iodide was prepared by reacting 65.0 mL of a solution containing 0.218 M lead(II) ions with 80.0 mL of a solution containing 0.265 M iodide ions. If the actual yield of the reaction was 3.26 g, which choice is closest to the %yield of the reaction?
and the choices are ....
is it 33.4%?
To find the percent yield of a reaction, we need to compare the actual yield (the obtained amount of product) with the theoretical yield (the maximum possible amount of product based on stoichiometry).
First, we need to determine the stoichiometry of the reaction. The balanced equation for the reaction is:
2Pb2+ + 4I- -> PbI2
From the equation, we can see that 2 moles of Pb2+ react with 4 moles of I- to produce 1 mole of PbI2.
Next, we calculate the amount of lead(II) iodide that should have formed based on the limiting reactant. To do this, we first determine which reactant is the limiting reactant by comparing the number of moles of each reactant present.
For lead(II) ions:
Volume = 65.0 mL
Concentration = 0.218 M
Number of moles = Volume * Concentration = 65.0 mL * 0.218 mol/L = 14.17 mmol
For iodide ions:
Volume = 80.0 mL
Concentration = 0.265 M
Number of moles = Volume * Concentration = 80.0 mL * 0.265 mol/L = 21.20 mmol
Since the stoichiometry of the reaction is 2:4, we can see that 2 moles of Pb2+ react with 4 moles of I-. Therefore, iodide ions are in excess, and lead(II) ions are the limiting reactant.
Based on stoichiometry, 2 moles of Pb2+ react to produce 1 mole of PbI2. Therefore, the number of moles of PbI2 formed is half of the moles of Pb2+.
Number of moles of PbI2 = 14.17 mmol / 2 = 7.08 mmol
Next, we need to calculate the theoretical yield of PbI2 using its molar mass. The molar mass of PbI2 is 461.01 g/mol.
Theoretical yield of PbI2 = Number of moles of PbI2 * Molar mass of PbI2
Theoretical yield of PbI2 = 7.08 mmol * 461.01 g/mol = 3260.04 mg = 3.26 g
The actual yield of the reaction is given as 3.26 g. To find the percent yield, we divide the actual yield by the theoretical yield and multiply by 100.
Percent yield = (Actual yield / Theoretical yield) * 100
Percent yield = (3.26 g / 3.26 g) * 100
Percent yield = 100%
Therefore, the closest choice to the percent yield of the reaction is 100%.