The correct option is C 1.5×10−3
Milli moles of Pb(NO3)2=35×1.5=52.5
Milli moles of KI=15×0.2=3.0Pb(NO3)2+2KI→PbI2+2KNO3
2 moles of KI react with 1 mole of Pb(NO3)2
moles of Pb(NO3)2 react with 3 mmoles of KI = 1×3×10−32=1.5×10−3
. Hence the limiting reagent is KI.
∴Moles of PbI2 formed =32×10−3=1.5×10−3