Determine the mass of the precipitate, lead iodide that is formed when 150 ml of 0.0500M pb(no3)2 is reacted with excess potassium iodide. if 3.3 of the precipitate is experimentally recovered, determine the percent yeild.
Pb(NO3)2 + 2KI ==> PbI2 + 2KNO3
mols Pb(NO3)2 = M x L = ?
Using the coefficients in the balanced equation, convert mols Pb(NO3)2 to mols PbI2 ppt.
Now convert mols PbI2 to grams PbI2. That's grams = mols x molar mass = ? That is the theoretical yield (TY). The actual yield (AY) is 3.3g.
% yield = (AY/TY)*100 = ?
To determine the mass of the precipitate, lead iodide, formed in the reaction, you need to follow the steps below:
1. Write down the balanced chemical equation for the reaction between Pb(NO3)2 and potassium iodide (KI).
Pb(NO3)2 + 2KI -> PbI2 + 2KNO3
2. Determine the limiting reagent, which is the reactant that will be completely consumed and will determine the amount of product formed. In this case, you are given that there is an excess of potassium iodide, so Pb(NO3)2 is the limiting reagent.
3. Convert the volume of Pb(NO3)2 solution into moles using the concentration (Molarity) given:
Molarity (M) = moles/volume (L)
Moles = Molarity * volume (in L)
Moles of Pb(NO3)2 = 0.0500 mol/L * 0.150 L = 0.0075 moles
4. Use the stoichiometry of the balanced equation to determine the moles of the precipitate formed. From the balanced equation, for every 1 mole of Pb(NO3)2, 1 mole of PbI2 is formed.
Moles of PbI2 = 0.0075 moles
5. Calculate the molar mass of PbI2. This can be found by adding the atomic masses of lead (Pb) and iodine (I) together.
Molar mass of PbI2 = atomic mass of Pb + 2 * atomic mass of I
= (207.2 g/mol) + 2 * (126.9 g/mol)
= 459 g/mol (approximately)
6. Calculate the mass of the precipitate using the moles of PbI2 and the molar mass:
Mass of PbI2 = moles of PbI2 * molar mass of PbI2
= 0.0075 moles * 459 g/mol
= 3.4425 grams (approximately)
So, the mass of the precipitate (lead iodide) formed is approximately 3.4425 grams.
To determine the percent yield, you need the experimental value (mass) of the recovered precipitate.
Percent yield = (experimental yield / theoretical yield) * 100
Given that 3.3 grams of the precipitate were experimentally recovered, the percent yield can be calculated as:
Percent yield = (3.3 g / 3.4425 g) * 100
= 95.8% (approximately)
Therefore, the percent yield of lead iodide is approximately 95.8%.