"How many grams of the precipitate will form if 25.5 mL of 4.5 M solution of
lead (II) nitrate is allowed to react with 35.5 mL of 3.0 M solution of
potassium chloride? What is the concentration of the excess reactant after the reaction has reached completion?"
I got that the limiting is KCl and the excess is PbCl2. How do I find the concentration after the experiment? I get 4.5M, but I feel that's not right.
that's what you start with
... some of it is precipitated out
also, the volume of the solution is more than doubled
(remaining PbCl2) / solution volume
millimoles = mmols.
Pb(NO3)2 + 2KCl ==> PbCl2 + 2KNO3
mmols Pb(NO3)2 = mL x M = 25.5 x 4.5 = 114.75
mmols PbCl2 formed if you had all of the KCl you needed = 114.25
mmols KCl = mL x M = 35.5 x 3.0 = 106.5
mmols PbCl2 if you had all of the Pb(NO3)2 you needed = 53.25.
So KCl is the limiting reagent, you will form 53.25 mmols (0.05325 mols) PbCl2 or g PbCl2 = mols PbCl2 x molar mass PbCl2 = ? g PbCl2.
So you use all of the KCl. How much Pb(NO3)2 is used? That's 53.5 mmols KCl x [1 mol Pb(NO3)2/2 mols KCl)] = 106.5 x 1/2 = 53.25 mmols Pb(NO3)2 used. How much Pb(NO3)2 is left? That's 114.75-53.25 = 61.5 mmols Pb(NO3)2 not used. Then you have mmols/mL = 61.5/total mL = 61.5/61.0 mL = M Pb(NO3)2 in solution which is the excess reagent and that's what the problem asked, not (PbCl2).
Check all of these figures.
To find the concentration of the excess reactant after the reaction has reached completion, you need to determine the limiting reactant and the amount of the excess reactant that remains.
1. Determine the limiting reactant:
To do this, compare the stoichiometry of the balanced chemical equation with the given amounts of the reactants. The balanced chemical equation for the reaction between lead (II) nitrate (Pb(NO3)2) and potassium chloride (KCl) is:
Pb(NO3)2 + 2KCl -> PbCl2 + 2KNO3
According to the equation, the molar ratio between lead (II) nitrate and potassium chloride is 1:2. Convert the given amounts of reactants to moles using their respective molarities:
Moles of Pb(NO3)2 = 25.5 mL * (4.5 moles/L) * (1 L / 1000 mL)
Moles of KCl = 35.5 mL * (3.0 moles/L) * (1 L / 1000 mL)
Calculate the mole ratio between the two reactants:
Moles of Pb(NO3)2 / 1 = Moles of KCl / 2
If the mole ratio is less than 1/2, then KCl is the limiting reactant. In this case, since the mole ratio is equal to 1/2, neither Pb(NO3)2 nor KCl is limiting and both are in excess.
2. Calculate the amount of excess reactant remaining:
Since both reactants are in excess, the concentration of the excess reactant can be found by subtracting the amount consumed from the initial amount. Since the amounts of both reactants are given in mL, you need to convert them to moles before performing the calculation.
Moles of KCl remaining = (Initial moles of KCl) - (moles of Pb(NO3)2 consumed)
Moles of KCl remaining = Moles of KCl - (Moles of Pb(NO3)2 * (2 mol of KCl / 1 mol of Pb(NO3)2))
3. Calculate the concentration of the excess reactant:
To find the concentration of the excess reactant, divide the moles of the excess reactant remaining by the total volume of the excess reactant, which is the sum of the initial volumes of both reactants.
Concentration of KCl remaining = (Moles of KCl remaining) / (Total volume of KCl)
Evaluate these calculations using the given values in the problem to find the concentration of the excess reactant after the reaction has reached completion.