If 150 mL of 1.73 M aqueous NaI and 235 g of Pb(NO3)2 are reacted stoichiometrically according to the balanced equation, how many grams of Pb(NO3)2 remain?

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

Molar Mass (g/mol)
NaI 149.89
Pb(NO3)2 331.21
Density (g/mL)
-
Molar Volume (L)
22.4 at STP
Gas Constant
(L.atm.mol-1.K-1)
0.0821


show steps please

To find the amount of Pb(NO3)2 remaining, you need to first determine the number of moles of NaI and Pb(NO3)2 initially present, and then use the stoichiometry of the balanced equation to calculate the amount of Pb(NO3)2 consumed. Finally, subtracting the amount of Pb(NO3)2 consumed from the initial amount will give you the remaining amount.

Step 1: Calculate the number of moles of NaI present.
Given:
Volume of NaI solution = 150 mL
Molarity of NaI solution = 1.73 M

Using the equation: Moles = Volume (L) x Molarity

Convert the volume from mL to L by dividing by 1000:
150 mL ÷ 1000 = 0.150 L

Moles of NaI = 0.150 L x 1.73 M = 0.2595 moles

Step 2: Calculate the number of moles of Pb(NO3)2 present.
Given:
Mass of Pb(NO3)2 = 235 g
Molar mass of Pb(NO3)2 = 331.21 g/mol

Moles of Pb(NO3)2 = Mass / Molar mass
Moles of Pb(NO3)2 = 235 g / 331.21 g/mol = 0.709 moles

Step 3: Determine the stoichiometric relationship between NaI and Pb(NO3)2 from the balanced equation.
The balanced equation tells us that 1 mole of Pb(NO3)2 reacts with 2 moles of NaI.

Step 4: Use stoichiometry to calculate the moles of Pb(NO3)2 consumed.
Since the reaction is stoichiometrically balanced, we know that the ratio of moles of Pb(NO3)2 consumed to NaI is 1:2.

Moles of Pb(NO3)2 consumed = (Moles of NaI) / 2
Moles of Pb(NO3)2 consumed = 0.2595 moles / 2 = 0.12975 moles

Step 5: Calculate the remaining moles of Pb(NO3)2.
Remaining moles of Pb(NO3)2 = Initial moles - Consumed moles
Remaining moles of Pb(NO3)2 = 0.709 moles - 0.12975 moles = 0.57925 moles

Step 6: Calculate the remaining mass of Pb(NO3)2.
Mass of Pb(NO3)2 = Remaining moles x Molar mass
Mass of Pb(NO3)2 = 0.57925 moles x 331.21 g/mol = 191.81 grams

Therefore, the remaining mass of Pb(NO3)2 is 191.81 grams.