Question

Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction

Pb(NO3)2(aq)+2NH4I(aq) -> PbI2(s) + 2NH4NO3(aq)

What volume of a 0.530 M NH4I solution is required to react with 155 mL of a 0.580 M Pb(NO3)2 solution?

How many moles of PbI2 are formed from this reaction?

Answer 1

Concentration of Pb(NO3)2 = 0.580 M = 0.580 mol/L

Volume of Pb(NO3)2 solution = 155 ml = 155 L/1000 = 0.155 L

Number of moles of Pb(NO3)2 = 0.580 mol/L * 0.155 L = 0.0899 mol

From reaction, number of moles of NH4I required = 2 * number of moles of Pb(NO3)2

So number of moles of NH4I required = 2 * 0.0899 mol = 0.1798 mol

Concentration of NH4I = 0.530 M = 0.530 mol/L

Volume of NH4I required = 0.1798 mol / 0.530 mol/L = 0.3392 L = 0.3392 * 1000 ml = 339.2 ml

From reaction, Number of moles of Pb(NO3)2 reacted = number of moles of PbI2 formed

So number of moles of PbI2 formed = 0.0899 mol