I don't know if i am solving for the number moles of NaI correctly, i multiply .21 by 1.67, is that incorrect? i know that to solve this you have to find the limiting reagent, but i think the problem might be that i am solving for moles wrong.

If 210 mL of 1.67 M aqueous NaI and 156 g of Pb(NO3)2 are reacted stoichiometrically according to the balanced equation, how many moles of solid PbI2 are produced?

Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq

moles NaI = M x L.

moles Pb(NO3)2 = grams/molar mass.

To solve for the number of moles of NaI correctly, you need to use the given information and the stoichiometry of the balanced equation.

First, let's calculate the number of moles of NaI present in 210 mL of 1.67 M NaI solution.

Step 1: Convert the volume from milliliters (mL) to liters (L).
210 mL * (1 L / 1000 mL) = 0.210 L

Step 2: Use the formula:
moles = concentration * volume
moles of NaI = 1.67 M * 0.210 L = 0.3507 moles of NaI

Now, to find the moles of solid PbI2 produced, we need to determine the limiting reactant. The limiting reactant is the one that will be completely consumed and determine the amount of product formed.

Step 3: Calculate the number of moles of Pb(NO3)2 using its molar mass.
Given mass of Pb(NO3)2 = 156 g
Molar mass of Pb(NO3)2 = (207 g/mol + 3 * (14 g/mol + 16 g/mol)) = 331 g/mol
moles of Pb(NO3)2 = 156 g / 331 g/mol ≈ 0.471 moles of Pb(NO3)2

Step 4: Use the stoichiometry of the balanced equation to determine the limiting reactant.
According to the balanced equation, the stoichiometric ratio between Pb(NO3)2 and NaI is 1:2. This means that for every mole of Pb(NO3)2, you need two moles of NaI.

We have:
moles of NaI = 0.3507 moles
moles of Pb(NO3)2 = 0.471 moles

Since the ratio is 1:2, the limiting reactant is the one that gives us the smaller number of moles when comparing to the stoichiometry ratio.

Here, 0.3507 moles of NaI is less than what is needed in the stoichiometry (1 mole of Pb(NO3)2 requires 2 moles of NaI). Therefore, NaI is the limiting reactant.

Step 5: Use the stoichiometry of the balanced equation to calculate the moles of PbI2 formed.
From the balanced equation, we see that the stoichiometry ratio between PbI2 and NaI is 1:2. This means that for every mole of PbI2, we need two moles of NaI.

Since NaI is the limiting reactant, we use its moles to find the moles of PbI2.
moles of PbI2 = 0.3507 moles of NaI * (1 mole of PbI2 / 2 moles of NaI) = 0.1754 moles of PbI2

So, the number of moles of solid PbI2 produced is 0.1754 moles.