When aqueous solutions of lead(II) ion are treated with potassium chromate solution, a bright yellow precipitate of lead(II) chromate, PbCrO4, forms. How many grams of lead chromate form when a 1.60-g sample of Pb(NO3)2 is added to 24.0 mL of 1.01 M K2CrO4 solution?

To determine the mass of lead chromate (PbCrO4) formed, we need to use stoichiometry. The balanced chemical equation for the reaction involving lead(II) ion (Pb2+) and potassium chromate (K2CrO4) is:

Pb(NO3)2 + K2CrO4 → PbCrO4 + 2 KNO3

From the equation, we can see that the mole ratio between Pb(NO3)2 and PbCrO4 is 1:1. We also need to convert the volume of K2CrO4 solution to moles using its molarity.

First, let's calculate the moles of Pb(NO3)2:

Number of moles of Pb(NO3)2 = Mass / Molar mass
Number of moles of Pb(NO3)2 = 1.60 g / (207.2 g/mol) (molar mass of Pb(NO3)2 is 207.2 g/mol)
Number of moles of Pb(NO3)2 = 0.00772 mol

Next, let's calculate the number of moles of K2CrO4:

Number of moles of K2CrO4 = Concentration × Volume
Number of moles of K2CrO4 = 1.01 M × 0.0240 L (volume converted to liters)
Number of moles of K2CrO4 = 0.0242 mol

Now, we can compare the moles of Pb(NO3)2 and K2CrO4 to determine the limiting reagent. The limiting reagent is the one with fewer moles.

The number of moles of Pb(NO3)2 is 0.00772 mol and the number of moles of K2CrO4 is 0.0242 mol. Since Pb(NO3)2 has fewer moles, it is the limiting reagent.

Using the stoichiometric ratio, we know that 1 mole of Pb(NO3)2 produces 1 mole of PbCrO4.

Therefore, the number of moles of PbCrO4 formed is also 0.00772 mol.

Finally, let's calculate the mass of PbCrO4 formed:

Mass of PbCrO4 = Number of moles × Molar mass
Mass of PbCrO4 = 0.00772 mol × (323.2 g/mol) (molar mass of PbCrO4 is 323.2 g/mol)
Mass of PbCrO4 = 2.50 g (rounded to two decimal places)

So, when a 1.60 g sample of Pb(NO3)2 is added to 24.0 mL of 1.01 M K2CrO4 solution, approximately 2.50 grams of lead chromate (PbCrO4) will form.

To determine the amount of lead chromate (PbCrO4) formed when a 1.60 g sample of Pb(NO3)2 is added to 24.0 mL of 1.01 M K2CrO4 solution, we'll use stoichiometry and the concept of limiting reactants.

First, let's write down the balanced chemical equation for the reaction:

Pb(NO3)2 + K2CrO4 → PbCrO4 + 2KNO3

From the equation, we can deduce that it takes a 1:1 ratio of Pb(NO3)2 to PbCrO4. Therefore, the molar amount of PbCrO4 formed will be equal to the molar amount of Pb(NO3)2.

Step 1: Calculate the molar amount of Pb(NO3)2
To do this, we will use the molar mass of Pb(NO3)2, which is 331.2 g/mol.

molar amount of Pb(NO3)2 = mass / molar mass
molar amount of Pb(NO3)2 = 1.60 g / 331.2 g/mol

Step 2: Calculate the volume of the K2CrO4 solution
To do this, we will use the concentration and volume information given.

volume of K2CrO4 solution = 24.0 mL = 0.0240 L

Step 3: Calculate the number of moles of K2CrO4
To do this, we will use the molarity (concentration) of the K2CrO4 solution.

moles of K2CrO4 = concentration × volume
moles of K2CrO4 = 1.01 M × 0.0240 L

Step 4: Identify the limiting reactant
To determine the limiting reactant, we compare the molar amounts of Pb(NO3)2 and K2CrO4. The reactant with the smaller molar amount is the limiting reactant.

The molar amount of Pb(NO3)2 is the same as calculated in step 1.
The molar amount of K2CrO4 is the same as calculated in step 3.

Step 5: Calculate the molar amount of PbCrO4 formed
Since Pb(NO3)2 and PbCrO4 have a 1:1 stoichiometric ratio, the molar amount of PbCrO4 formed will be the same as the molar amount of Pb(NO3)2.

molar amount of PbCrO4 = molar amount of Pb(NO3)2

Step 6: Calculate the mass of PbCrO4 formed
To do this, we will use the molar mass of PbCrO4, which is 323.2 g/mol.

mass of PbCrO4 = molar amount × molar mass
mass of PbCrO4 = molar amount of PbCrO4 × 323.2 g/mol

Now, you can substitute the values calculated in each step to find the mass of PbCrO4 formed.

I don't see a Ksp given for PbCrO4; therefore, I will assume this is just another limiting reagent problem. I work these the long way.

mols Pb(NO3)2 = grams/molar mass
mols K3CrO4 = M x L = ?

Using the coefficients in the balanced equation, convert mols Pb(NO3)2 to mols PbCrO4.
Do the same and convert mols K2CrO4 to mols PbCrO4.
It is likely that the two values will not be the same; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent responsible for that is the limiting reagent.
Using the smaller number of mols of PbCrO4, convert to grams. g = mols x molar mass.