Suppose that 3.50 g of solid potassium chromate is added to 75.0 mL and 0.250 M Mg(NO3)2(aq).
a) What is the initial molarity of potassium chromate in the solution? b) What mass of potassium
is present in solution? Write the formula for the precipitate that forms. Assume a final volume
of 75.0 mL.
The equation of reaction is next:
K2CrO4 + Mg(NO3)2 = MgCrO4 (precipitate) + 2KNO3
The amount of Mg(NO3)2 is : 75.0*0.250/1000 = 0.01875 mol The amount of K2CrO4 = 3.5/194 = 0.01804 mol
Mg(NO3)2 is in excess and it means that there is no K2CrO4 in solution after and initial molarity of potassium chromate in the solution = 0
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