An aqueous solution containing 9.23 g of lead(II) nitrate is added to an aqueous solution containing 7.20 g of potassium chloride.
The percent yield for the reaction is 86.6%. How many grams of the precipitate are formed and how many grams of the excess reactant remain?
To determine the grams of precipitate formed and the grams of excess reactant remaining, we need to consider the balanced equation for the reaction between lead(II) nitrate (Pb(NO3)2) and potassium chloride (KCl):
Pb(NO3)2 + 2KCl -> PbCl2 + 2KNO3
From the balanced equation, we can see that the stoichiometric ratio between Pb(NO3)2 and PbCl2 is 1:1. This means that for every 1 mole of Pb(NO3)2, we will get 1 mole of PbCl2.
First, let's calculate the moles of lead(II) nitrate (Pb(NO3)2) and potassium chloride (KCl) in the given amounts:
Molar mass of Pb(NO3)2 = 207.2 g/mol
Molar mass of KCl = 74.6 g/mol
Moles of Pb(NO3)2 = mass / molar mass = 9.23 g / 207.2 g/mol ≈ 0.0445 mol
Moles of KCl = mass / molar mass = 7.20 g / 74.6 g/mol ≈ 0.0966 mol
Next, we need to determine the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of product formed. To find the limiting reactant, we compare the moles of Pb(NO3)2 and KCl using their stoichiometric ratio:
Moles of Pb(NO3)2 / stoichiometric coefficient of Pb(NO3)2 = 0.0445 mol / 1 = 0.0445 mol
Moles of KCl / stoichiometric coefficient of KCl = 0.0966 mol / 2 = 0.0483 mol
The stoichiometric coefficient of Pb(NO3)2 is 1, and the stoichiometric coefficient of KCl is 2 in the balanced equation. Since the moles of Pb(NO3)2 (0.0445 mol) are lower than the moles of KCl (0.0483 mol), Pb(NO3)2 is the limiting reactant.
Now, let's calculate the theoretical yield of PbCl2 using the stoichiometric ratio:
Moles of Pb(NO3)2 = Moles of PbCl2 (according to stoichiometry)
The theoretical yield of PbCl2 is equal to the moles of Pb(NO3)2, which is 0.0445 mol.
To calculate the grams of the precipitate (PbCl2) formed, we can multiply the moles of PbCl2 by its molar mass:
Molar mass of PbCl2 = 278.1 g/mol
Grams of PbCl2 formed = moles of PbCl2 × molar mass of PbCl2
= 0.0445 mol × 278.1 g/mol
≈ 12.37 g
The percent yield is given as 86.6%, so we can calculate the actual yield of PbCl2:
Actual yield of PbCl2 = percent yield × theoretical yield
= 0.866 × 12.37 g
≈ 10.69 g
Finally, to determine the grams of excess reactant remaining, we can subtract the moles of the limiting reactant from the moles of the excess reactant:
Moles of excess KCl = moles of KCl - moles of Pb(NO3)2
= 0.0966 mol - 0.0445 mol
≈ 0.0521 mol
To calculate the grams of excess KCl remaining, we can multiply the moles of excess KCl by its molar mass:
Grams of excess KCl remaining = moles of excess KCl × molar mass of KCl
= 0.0521 mol × 74.6 g/mol
≈ 3.88 g
Therefore, approximately 12.37 grams of PbCl2 are formed as a precipitate, and approximately 3.88 grams of excess KCl remain.