Posted by darren on Tuesday, November 29, 2011 at 4:33pm.
K2CrO4 + 2AgNO3 ==> Ag2CrO4 + 2KNO3
moles K2CrO4 = M x L = ?
moles AgNO3 = M x L = ?
Now do two stoichiometry problems.
1. How many moles Ag2CrO4 if we used all of the K2CrO4. moles Ag2CrO4 = moles K2CrO4 x (1 mole Ag2CrO4/1 mole K2CrO4) = moles K2CrO4 x (1/1) = ?
2. How many moles Ag2CrO4 if we used all of the AgNO3? moles Ag2CrO4 = moles AgNO3 x (1 mole Ag2CrO4/2 moles AgNO3) = moles AgNO3 x (1/2) = ?
Calculate 1 gives you a different value than calculation 2. Both answers can't be right; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Use the smaller value, convert to grams. g = moles x molar mass.
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