Question

Consider the mixing of 3.05 mL of 0.220 M Mg(NO₃)₂ and 1.00 L of 0.180 M K₃PO₄.

(a) Write a balanced chemical equation that describes the double displacement reaction between these solutions. (Include states-of-matter under the given conditions in your answer.)

(b) Calculate how many grams of the precipitate will form.

What would be the concentration of a solution made by diluting 46.9 mL of 4.54 M KOH to 625 mL?

Answer 1

a)

balanced chemical equation

3 Mg(NO3)2 (aq) + 2 K3PO4 (aq) ----------------------------> Mg3(PO4)2 (s) +6 KNO3 (aq)

b)

moles of Mg(NO3)2 = 3.05 x 0.220 / 1000 = 6.71 x 10^-4

moles of K3PO4 = 1 x 0.180 = 0.180

3 Mg(NO3)2 (aq) + 2 K3PO4 (aq) ----------------------------> Mg3(PO4)2 (s) +6 KNO3 (aq)

3                                 1

6.71 x 10^-4           0.180

limiting reagent is Mg(NO3)2

so moles of precipitate formed = 6.71 x 10^-4 / 3 = 2.24 x 10^-4

mass of precipitate = 2.24 x 10^-4 x 262.9

                               = 0.0588 g

mass of precipitate = 0.0588 g

2)

concentration = 46.9 x 4.54 / 625

                       = 0.341 M

concentration = 0.341 M