If 740. mL of 0.06622 M aqueous NaI and 370. mL of 0.07151 M aqueous Pb(NO3)2 are reacted stoichiometrically according to the equation, what mass (g) of NaI remained?

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

See above.

To find the mass of NaI that remained, we first need to determine which reactant is limiting. This can be done by comparing the number of moles of each reactant, using the given volumes and concentrations.

Step 1: Calculate the number of moles of NaI and Pb(NO3)2:

Given:
Volume of NaI solution = 740 mL
Concentration of NaI solution = 0.06622 M

Moles of NaI = Concentration × Volume
Moles of NaI = 0.06622 M × 0.740 L (since 1 L = 1000 mL)
Moles of NaI = 0.04891 moles

Given:
Volume of Pb(NO3)2 solution = 370 mL
Concentration of Pb(NO3)2 solution = 0.07151 M

Moles of Pb(NO3)2 = Concentration × Volume
Moles of Pb(NO3)2 = 0.07151 M × 0.370 L
Moles of Pb(NO3)2 = 0.02647 moles

Step 2: Determine the stoichiometric ratio between NaI and Pb(NO3)2:

From the balanced equation, we can see that the stoichiometric ratio between NaI and Pb(NO3)2 is 2:1. This means that for every 2 moles of NaI, we need 1 mole of Pb(NO3)2.

Step 3: Calculate the number of moles of NaI that would be needed to react with all the Pb(NO3)2:

Since the stoichiometric ratio is 2:1, the number of moles of NaI needed is half the number of moles of Pb(NO3)2.

Moles of NaI needed = 0.02647 moles of Pb(NO3)2 / 2
Moles of NaI needed = 0.01324 moles

Step 4: Calculate the mass of NaI that remained:

Molar mass of NaI = (22.99 g/mol) + (126.9 g/mol)
Molar mass of NaI = 149.89 g/mol

Mass of NaI (grams) = Moles of NaI remaining × Molar mass of NaI
Mass of NaI (grams) = (0.04891 moles - 0.01324 moles) × 149.89 g/mol
Mass of NaI (grams) = 5.863 grams (rounded to three decimal places)

Therefore, the mass of NaI that remained is approximately 5.863 grams.