You have 75.0 mL of 2.50 M solution of Na2CrO4 (aq). You also have 125 mL...

You have 75.0 mL of 2.50 M solution of Na2CrO4 (aq). You also have 125 mL of a 2.01 M solution of AgNO3 (aq). Calculate the mass of Ag2CrO4 precipitate that is produced after the two solutions are mixed together and reaction has occured.

Expert Solution

Na2CrO4 + 2AgNO3 ----> Ag2CrO4 + 2NaNO3
no of moles of Na2CrO4 = molarity * volume in L
                       = 2.5*0.075
                       = 0.1875 moles
no of moles of AgNO3   = molarity * volume in L

                       = 2.01*0.125
                       = 0.25125 moles
AgNo3 is limiting reagent ( 2*0.1875 = 0.375 required)
2 moles of AgNo3 react with Na2CrO4 to gives1 mole of Ag2Cro4
0.25125 moles of AgNo3 reat with na2CrO4 to gives = 0.25125*1/2
                                                  =0.125625 moles of Ag2CrO4
mass of Ag2CrO4 = no of moles * gram molar mass
                = 0.125625*331.73
mass of Ag2Cro4 = 41.67gm >>>>> answer

queen_honey_blossom answered 3 years ago

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