Final answer:
To completely react with 683 mL of a 0.300 M Pb(NO3)2 solution, 2.73 L of a 0.150 M NH4I solution is required. The reaction will produce 0.205 moles of PbI2.
Step-by-step explanation:
To determine the volume of the NH4I solution required to react with the Pb(NO3)2 solution, we need to use the balanced equation:
Pb(NO3)2(aq) + 2NH4I(aq) → PbI2(s) + 2NH4NO3(aq)
The molar ratio between Pb(NO3)2 and NH4I is 1:2, meaning that 1 mole of Pb(NO3)2 reacts with 2 moles of NH4I. First, we calculate the number of moles of Pb(NO3)2:
Moles of Pb(NO3)2 = Molarity x Volume(L)
Moles of Pb(NO3)2 = 0.300 mol/L x 0.683 L = 0.205 moles
Since the molar ratio is 1:2, the number of moles of NH4I required is twice that of Pb(NO3)2:
Number of moles of NH4I = 2 x 0.205 moles = 0.410 moles
To calculate the volume of the NH4I solution, we rearrange the formula:
Volume(L) = Moles of NH4I / Molarity
Volume(L) = 0.410 moles / 0.150 mol/L = 2.73 L
Therefore, 2.73 L of the 0.150 M NH4I solution are required to react with 683 mL of the 0.300 M Pb(NO3)2 solution.
To determine the number of moles of PbI2 formed, we use the molar ratio between Pb(NO3)2 and PbI2, which is 1:1. Since we calculated the moles of Pb(NO3)2 to be 0.205 moles, the moles of PbI2 formed will also be 0.205 moles.