If 360 mL of 2.47 M aqueous NaI and 212 g of Pb(NO3)2 are reacted stoichiometrically according to the balanced equation, how many milliliters of 4.62 M aqueous NaNO3 are produced?

Pb(NO3)2(aq) + 2NaI(aq) �¨ PbI2(s) + 2NaNO3(aq)

i figured it out...nvm

To solve this question, you'll need to use stoichiometry to determine the number of moles of each reactant and product involved in the reaction. Then, you can use the molarity and volume information given to calculate the volume of aqueous NaNO3 produced.

First, let's find the number of moles of NaI and Pb(NO3)2. We can calculate this using the given volume and molarity of the NaI solution.

Number of moles of NaI = Volume (in L) × Molarity = 360 mL × (1 L/1000 mL) × 2.47 mol/L

Similarly, let's find the number of moles of Pb(NO3)2 using its mass.

Number of moles of Pb(NO3)2 = Mass (in grams) / Molar mass
Number of moles of Pb(NO3)2 = 212 g / (207.2 g/mol + 3 × 16.0 g/mol)

Now, we can use the stoichiometry of the balanced equation to determine the ratio of moles of NaNO3 produced to moles of Pb(NO3)2 consumed. From the balanced equation, we see that 1 mole of Pb(NO3)2 reacts to form 2 moles of NaNO3.

Next, we can calculate the number of moles of NaNO3 produced by multiplying the number of moles of Pb(NO3)2 by the mole ratio:

Number of moles of NaNO3 = Number of moles of Pb(NO3)2 × (2 moles NaNO3 / 1 mole Pb(NO3)2)

Finally, we can calculate the volume of 4.62 M aqueous NaNO3 produced using the molarity and the number of moles of NaNO3:

Volume (in L) = Number of moles of NaNO3 / Molarity = Number of moles of NaNO3 / 4.62 mol/L

To convert the volume to milliliters, simply multiply by 1000:

Volume (in mL) = Volume (in L) × 1000

Now you have the steps to calculate the volume of 4.62 M aqueous NaNO3 produced in this reaction.