# chem

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mix 35.50 mL of a 0.201 M solution of K2CrO4 with 35.50 mL of a 0.201 M solution of AgNO3, what mass of solid will form?

• chem -

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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