mols Pb(NO3)2 = 9.23/331.2 = 0.0279
mols KCl = 7.2/74.6 = .0965
..........Pb(NO3)2 + 2KCl ==> PbCl2(s) + 2KNO3
I..........0.0279........0.0965........0..................0
C........-0.0279......-2*0.0279 ....0.0279.......
E.............0............-0.0407........0.0279
If this is a lower level chemistry class, and I assume that is so, then 0.0407 mols KCl remain and that is the excess reagent. Convert to grams by grams = mols x molar mass = ?
An aqueous solution containing 9.23 g of lead(II) nitrate is added to an aqueous solution containing 7.20 g of potassium chloride.
How many grams of the excess reactant remain?
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