Question

Assume the reaction goes to completion, and the volumes are additive, when 35.0 ml of 0.25M tin(II) nitrate is combined with 45.0 ml of 0.30 M potassium phosphate.

a. calculate the mass of precipitate that should form.

b. In the lab, the precipitate was washed, dried and massed as 1.00g, calculate the % yield of ppt formed.

c. Determine the concentration (molarity) of each ion still floating in the solution.

Answer 1

concnetration of Sn(NO3)2 = 35 x 0.25 / (35 + 45) = 0.109 M

concentration of K3PO4 = 45 x 0.30 / 35 + 45 = 0.169 M

a)

3 Sn(NO3)2   +   2 K3PO4    ----------------> Sn3(PO4)2 (s) + 6 KNO3

      3                       2                                            1

0.00875             0.0135                                        ??

here limitng reagent is Sn(NO3)2.

3 mol Sn(NO3)2 ----------> 1 mol Sn3(PO4)2

0.00875 mol     --------------> ??

moles of Sn3(PO4)2 = 0.00875 x 1 / 3 = 2.92 x 10^-3

mass of Sn3(PO4)2 = 1.59 g

b)

actual mass = 1.00 g

% yield = (actual / theoretical ) x 100

            = 1.00 / 1.59 ) x 100

% yield = 62.9 %

c)

[NO3-] = 0.218 M

[PO43-] = 0.0963 M

[K+] = 0.289 M