Asked by Help me
                60 mL of 0.322 M potassium iodide are combined with 20.0 mL of 0.530 m lead(II) nitrate. How many grams of lead (II) iodide will precipitate?
            
            
        Answers
                    Answered by
            oobleck
            
    2KI + Pb(NO3)2 = PbI2 + 2KNO3
You will get half as many moles of PbI2 as you use moles of KI
so now just see which gets used up first, and that will tell you how many grams are produced.
    
You will get half as many moles of PbI2 as you use moles of KI
so now just see which gets used up first, and that will tell you how many grams are produced.
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