Answer to Question #20619 in Chemistry for jasie

Question

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.

Answer 1

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