Question

Complete a reaction table in millimoles for the reaction of 73.9 mL of 1.0200 M Ag1+ and 59.8 mL of 1.1200 M CrO42-.

2Ag1+ + CrO42- Ag2CrO4

initial mmol

delta mmol

final mmol

What mass in grams of Ag2CrO4 forms? g What is the molar concentration of the remaining excess reactant? Assume that the volumes are additive.

Answer 1

2Ag1+ + CrO42- ---> Ag2CrO4

initial

no of mole of Ag^+ = 73.9*1.02 = 75.38 mmol

no of mole of CrO4^2- = 59.8*1.12 = 66.976 mmol

limiting reactant = Ag^+

no of mmole of components changes(D mmol) = 66.976 mmol

no of mole of Ag2CrO4 = 75.38/2 = 37.69 mmol

mass of Ag2crO4 = 37.69*10^-3*331.73 = 12.52 g

remaining excess reagent = CrO4^2-

                          = 66.976 - (75.38/2) = 29.286 mmol

concentration of excess reagent = 29.286/(73.9+59.8)

                                 = 0.22 M