Write the balanced equation for the reaction of aqueous Pb(ClO3)2 with aqueous NaI. Include phases.

What mass of precipitate will form if 1.50 L of highly concentrated Pb(ClO3)2 is mixed with 0.350 L of 0.290 M NaI? Assume the reaction goes to completion.

The balanced equation for the reaction between aqueous Pb(ClO3)2 and aqueous NaI is:

Pb(ClO3)2(aq) + 2NaI(aq) -> PbI2(s) + 2NaClO3(aq)

To determine the mass of precipitate that will form, we need to use stoichiometry and the information given.

From the balanced equation, we can see that the mole ratio between Pb(ClO3)2 and PbI2 is 1:1. This means that for every 1 mole of Pb(ClO3)2 reacted, 1 mole of PbI2 is formed.

First, let's find the number of moles of Pb(ClO3)2 and NaI using the given volumes and concentration:

Pb(ClO3)2:
Volume = 1.50 L
Concentration is not given, assume highly concentrated solution means that it is 100% Pb(ClO3)2
Moles = Volume x Concentration = 1.50 L x 1000 g/L = 1500 g

NaI:
Volume = 0.350 L
Concentration = 0.290 M
Moles = Volume x Concentration = 0.350 L x 0.290 mol/L = 0.1015 moles

Since the balanced equation shows a stoichiometric ratio of 1:1 between Pb(ClO3)2 and PbI2, we can conclude that 0.1015 moles of Pb(ClO3)2 will produce 0.1015 moles of PbI2.

To find the mass of PbI2 formed, we need to multiply the number of moles by the molar mass of PbI2. The molar mass of PbI2 is 461 g/mol.

Mass of PbI2 = Moles of PbI2 x Molar mass of PbI2
Mass of PbI2 = 0.1015 moles x 461 g/mol
Mass of PbI2 = 46.5285 g

Therefore, the mass of precipitate (PbI2) that will form is approximately 46.53 grams.

To write the balanced equation for the reaction between aqueous Pb(ClO3)2 and aqueous NaI, we need to identify the individual ions present in each compound.

Pb(ClO3)2 dissociates into Pb2+ and ClO3- ions when dissolved in water.
NaI dissociates into Na+ and I- ions when dissolved in water.

So, when aqueous Pb(ClO3)2 reacts with aqueous NaI, the balanced equation is:

Pb(ClO3)2 + 2NaI -> PbI2 + 2NaClO3

Now, let's move on to calculating the mass of precipitate that will form.

First, we need to determine the limiting reagent. In this case, we compare the moles of Pb(ClO3)2 and NaI.

Given:
Volume of Pb(ClO3)2 = 1.50 L
Concentration of NaI = 0.290 M
Volume of NaI = 0.350 L

To calculate the moles of Pb(ClO3)2, we need to know the concentration of the solution.

Next, we can calculate the moles of NaI using the given concentration and volume:

moles of NaI = concentration of NaI × volume of NaI
= 0.290 M × 0.350 L

Now, compare the moles of Pb(ClO3)2 and NaI. Remember, the balanced equation shows a 1:2 molar ratio of Pb(ClO3)2 to NaI.

If the moles of Pb(ClO3)2 are more than twice the moles of NaI, NaI is the limiting reagent.

To find the mass of precipitate (PbI2) formed, we need to know the molar mass of PbI2. The molar mass of PbI2 is calculated by adding the atomic masses of lead (Pb) and iodine (I) with appropriate coefficients from the balanced equation.

Finally, multiplying the moles of PbI2 by its molar mass will give us the mass of precipitate formed.

Pb(ClO3)2 + 2NaI ==> PbI2 + 2NaClO3

mols NaI = M x L = ?
mols PbI2 = 1/2 mols NaI (use the coefficients in the balanced equation to see this)
g PbI2 = mols PbI2 x molar mass PbI2.