Mass → Mass Conversions

• Mass to Mass Problems involve one additional conversion

• Lead (IV) Nitrate Reacts with 5.0g of Potassium Iodide. How Many grams of Lead(IV) Nitrate are required for a complete reaction?

Pb(NO₃)₄ + 4KI → PbI₄ + 4 K(NO₃)
5.0g KI x I mol KI/ 166g KI = 0.030 mol KI
0.030 mol KI x Pb(NO₃)₄/ 4KI -0.0075 Pb(NO₃)₄
0.075 Pb(NO₃)₄ x  455.0 g Pb(NO3)4/  1 mol Pb(NO3)4 = 3.414g →3.4 g