Answer:
76.065 g of PbI2.
Explanation:
Equation of the reaction:
Pb(CH3CO2)2(aq) + 2NH4I --> PbI2(s) + 2NH4(CH3CO2)(aq)
Given:
Pb(CH3CO2)2 = 0.165 M
NH4I = 0.466 M
Since molar concentration = number of moles/volume
Calculating the number of moles of the limiting reagent,
0.165 moles of Pb(CH3CO2)2 in 0.466 moles of NH4I all in the same volume of solution and by stoichiometry, since 1 mole of Pb(CH3CO2)2 reacts with 2 moles of NH4I.
Therefore, Pb(CH3CO2)2 is the limiting reagent.
Using stoichiometry, 1 mole of Pb(CH3CO2)2 reacted to form 1 mole of PbI2 precipitate.
Number of moles of PbI2 = 0.165 moles
Molar mass of PbI2 = 207 + (127 * 2)
= 461 g/mol
Mass = molar mass * number of moles
= 461 * 0.165
= 76.065 g of PbI2.
