posted by strid on .
when 0.50 L of a 1.0 M solution of Pb (No3)2 is mixed with 0.50 L of a 1.0 M solution of KI, how many moles of lead (III) iodide are formed?
balance the equation.
Pb(NO3)2 + 2KI >> 2KNO3 + PbI2
You start with .5 mol lead nitrate, and .5 molk KI. So you don't have enough KI, it is the limiting reactant. If you use the .5 mol KI, you get .25 mol PbI2. Examine the coefficents to verify that.