Here's my working:
2KI + Pb(NO3)2 --> 2KNO3 + PBI2
n(KI) = Cv
= 1*0.023 mol
n(Pb(NO3)2) = 0.077 mol
Limiting reagant is the potassium iodide and from the formula, you need half as many mols of lead nitrate i.e. if we're using 0.023mols of KI, we're gonna need 0.0115 mols of...