0.04 moles of iodide is required to precipitate all the lead II ions from 25.0 mL of a 1.6 M Pb(NO3)2 solution.
The reaction equation is;
2KI(aq) + Pb(NO3)2(aq) -------> PbI2(s) + 2KNO3(aq)
The net ionic equation is;
Pb^2+(aq) + 2I^-(aq) ------> PbI2(s)
Number of moles of Pb(NO3)2 = 25/1000 L × 1.6 M = 0.04 moles
Number of moles of Pb^2+ = 0.04 moles /2 = 0.02 moles
Since 2 moles of iodide reacts with 1 mole of Pb^2+
x moles of iodide reacts with 0.02 moles of Pb^2+
x = 2 moles × 0.02 moles/ 1 mole
x = 0.04 moles of iodide
Hence, 0.04 moles of iodide is required to precipitate all the lead II ions from 25.0 mL of a 1.6 M Pb(NO3)2 solution.
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