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.

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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