The volume of potassium iodide used will be 0.087 L or 87 mL
Stoichiometric calculation
The equation of the reaction goes thus: [tex]Pb(NO_3)_2(aq) + 2KI --- > PbI_2(s) + 2KNO_3(aq)[/tex]
The mole ratio of the 2 reactants is 1:2.
Mole of 9.42 g [tex]Pb(NO_3)_2[/tex] = 9.42/331.2 = 0.02844 mol
Equivalent mole of KI = 0.02844 x 2 = 0.05688 mol
Volume of 0.05688 mol, 0.655 mol/L KI = 0.05688/0.655 = 0.087 L or 87 mL
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