I would be very grateful to have these answered with showing work
In order to produce a lead (II) chromate precipitate, lead (II) chloride reacts with sodium chromate in solution. A 12.5 g mass of lead (II) chloride is mixed into solution, and is allowed to react with excess sodium chromate.
a) what is the predicted yield of lead (II) chromate?
b) calculate the percentage yield if 13.8 g of lead (II) chromate is produced experimentally.
Well, you can start with
Na2CrO4 + Pb(NO3)2 -> PbCrO4 + NaNO3
Balance that equation, then convert the grams to moles to see what limits the reaction.
I should point out that PbCl2 is insoluble and normally one doesn't have solutions of lead(II) chloride.
To answer these questions, we need to balance the equation for the reaction between lead (II) chloride (PbCl2) and sodium chromate (Na2CrO4):
PbCl2 + Na2CrO4 -> PbCrO4 + 2NaCl
The balanced equation shows that 1 mole of lead (II) chloride reacts with 1 mole of sodium chromate to produce 1 mole of lead (II) chromate.
a) To calculate the predicted yield of lead (II) chromate, we need to convert the mass of lead (II) chloride to moles, and then use stoichiometry to determine the moles of lead (II) chromate produced.
The molar mass of PbCl2 is 207.2 g/mol.
12.5 g PbCl2 * (1 mol PbCl2 / 207.2 g PbCl2) = 0.0603 mol PbCl2
According to the balanced equation, 1 mole of PbCl2 produces 1 mole of PbCrO4.
Therefore, the predicted yield of lead (II) chromate is 0.0603 mol.
b) To calculate the percentage yield, we need to compare the actual yield (experimental yield) to the predicted yield (theoretical yield).
Actual Yield: 13.8 g PbCrO4
First, convert the actual yield to moles:
13.8 g PbCrO4 * (1 mol PbCrO4 / 323.2 g PbCrO4) = 0.0427 mol PbCrO4
Percentage Yield = (Actual Yield / Theoretical Yield) * 100
Percentage Yield = (0.0427 mol / 0.0603 mol) * 100 = 70.7%
Therefore, the percentage yield of lead (II) chromate is 70.7%.
To answer these questions, we need to use stoichiometry and the concept of limiting reagents. Let's go step by step.
a) To find the predicted yield of lead (II) chromate, we need to determine the limiting reagent in the reaction. The limiting reagent is the reactant that gets completely consumed and determines the maximum amount of product that can be formed.
First, we need to write and balance the chemical equation for the reaction:
PbCl2 + Na2CrO4 → PbCrO4 + 2NaCl
From the balanced equation, we see that one mole of lead (II) chloride (PbCl2) reacts with one mole of sodium chromate (Na2CrO4) to produce one mole of lead (II) chromate (PbCrO4).
Next, let's calculate the number of moles of lead (II) chloride (PbCl2) using its molar mass. The molar mass of PbCl2 is 207.2 g/mol.
Number of moles of PbCl2 = Mass of PbCl2 / Molar mass of PbCl2
= 12.5 g / 207.2 g/mol
≈ 0.0602 mol
Since the reaction is carried out with excess sodium chromate, it means that sodium chromate is not the limiting reagent. So, we will compare the number of moles of PbCl2 to the stoichiometric ratio of PbCl2 and PbCrO4.
From the balanced equation, we know that the stoichiometric ratio of PbCl2 to PbCrO4 is 1:1.
Therefore, the predicted yield of lead (II) chromate is also 0.0602 moles.
Finally, to convert the moles into grams, we can use the molar mass of lead (II) chromate (PbCrO4), which is 323.2 g/mol.
Predicted yield of lead (II) chromate = Moles of PbCrO4 * Molar mass of PbCrO4
≈ 0.0602 mol * 323.2 g/mol
≈ 19.47 g
So, the predicted yield of lead (II) chromate is approximately 19.47 grams.
b) To calculate the percentage yield, we need to compare the actual yield (experimental yield) with the predicted yield and then calculate the percentage.
Actual yield = 13.8 g
Predicted yield = 19.47 g
Now, we can use the formula for percentage yield:
Percentage yield = (Actual yield / Predicted yield) * 100
Percentage yield = (13.8 g / 19.47 g) * 100
≈ 70.99 %
Therefore, the percentage yield of lead (II) chromate is approximately 70.99%.