Question #20619

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.

Expert's answer

The equation of reaction is next:

K_{2}CrO_{4} + Mg(NO_{3})_{2} = MgCrO_{4} (precipitate) + 2KNO_{3}

The amount of Mg(NO_{3})_{2} is : 75.0*0.250/1000 = 0.01875 mol

The amount of K_{2}CrO_{4} = 3.5/194 = 0.01804 mol

Mg(NO_{3})_{2} is in excess and it means that there is no K_{2}CrO_{4} in solution after and initial molarity of potassium chromate in the solution = 0

K

The amount of Mg(NO

The amount of K

Mg(NO

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