Write the balanced equation for (aq) Pb(ClO3)2 and (aq) NaI. Include phases.

What mass of precipitate will form if 1.50 L of concentrated Pb(ClO3)2 is mixed with .250 L of .110 M NaI? Assume the reaction goes to completion.

This is a limiting reagent (LR) problem and you solve these by essentially working two simple stoichiometry problems like the Zn/HCl problem below.

Pb(ClO3)2 + 2NaI ==> PbI2(s) + 2NaClO3
mols Pb(ClO3) = M x L = ?
mols NaI = M x L = ?

Now using the coefficients in the balanced equation, convert mols Pb(ClO3)2 to mols PbI2.
Do the same for mols NaI to mols PbI2.
It is likely that these two values for mols PbI2 will not agree which means one of them is wrong. The correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
Now use the smaller value of PbI2 and convert it to grams. g = mols x molar mass.

You see, you just work two of the simple problems [Pb(ClO3)2 to PbI2 and NaI to PbI2], take the smaller value which identifies the LR, then convert that to grams just as before.

Are the moles of Pb(ClO3)2 .01375?

I don't know. I assumed you knew M Pb(ClO3)2 from the phrase "concentrated solution Pb(ClO3)2".

It's hard for me to wrap my brain around. I just learned all of this stuff! AH

6.34 g??

Yes, I would go with 6.34. What we are doing is assume a concentrated solution is enough to make it the excess reagent, then work the simple stoichiometric problem.

To write a balanced equation for the reaction between (aq) Pb(ClO3)2 and (aq) NaI, we need to identify the products formed. In this case, a double replacement reaction occurs, resulting in the formation of (s) PbI2 and (aq) NaClO3.

The balanced equation for the reaction is:
Pb(ClO3)2(aq) + 2 NaI(aq) → PbI2(s) + 2 NaClO3(aq).

Now, let's calculate the mass of the precipitate formed when 1.50 L of concentrated Pb(ClO3)2 is mixed with 0.250 L of 0.110 M NaI.

First, we need to determine the number of moles of Pb(ClO3)2 and NaI using their respective concentrations and volumes:

For Pb(ClO3)2:
Concentration (C1) = concentrated Pb(ClO3)2 = concentration in mol/L
Volume (V1) = 1.50 L
Number of moles (n1) = C1 × V1

For NaI:
Concentration (C2) = 0.110 M
Volume (V2) = 0.250 L
Number of moles (n2) = C2 × V2

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 number of moles of Pb(ClO3)2 and NaI. The stoichiometry of the balanced equation tells us that 1 mole of Pb(ClO3)2 reacts with 2 moles of NaI.

If the mole ratio is such that the moles of NaI is less than half the moles of Pb(ClO3)2, NaI is the limiting reactant. If NaI is greater than half the moles of Pb(ClO3)2, Pb(ClO3)2 is the limiting reactant.

Next, we calculate the number of moles of the precipitate formed using the limiting reactant:

For PbI2:
Number of moles (n3) = (0.5 × n1) (since the stoichiometry is 1:2)
Molar mass of PbI2 (M) = molar mass of element Pb + 2 × molar mass of element I

Finally, we determine the mass of the precipitate formed using the number of moles and the molar mass:

Mass of PbI2 = n3 × M

Note: The answer may need to be rounded to the appropriate significant figures based on the given data and the molar masses used.

By following these steps, you should be able to determine the balanced equation and calculate the mass of the precipitate formed in the given reaction.