PLEASE HELP ASAP

Question

The precipitate of silver fosphate is obtained by mixing 35.0 mL of 0.428 M sodium fosphate (Na3PO4) and 15.0 mL of 0.167 M silver nitrate (AgNO3).

a)How many grams of silver fosphate is obtained?

b) What is the concentration of excess amount of substance after the precipitaion.

Please write the solution with full steps

Answer 1

First, write a correctly balanced equation for the reaction takin place:

Na₃PO₄(aq) + 3AgNO₃(aq) ==> 3NaNO₃(aq) + Ag₃PO₄(s)

Next, since we are given the amounts of BOTH reactants, we must determine if either of them is limiting:

moles Na₃PO₄ present = 35.0 ml x 1 L/1000 ml x 0.428 mol/L = 0.01498 moles Na₃PO₄

moles AgNO₃ present = 15.0 ml x 1 L/1000 ml x 0.167 mol/L = 0.002505 moles AgNO₃

Since it requires 3 mol AgNO₃ per 1 mol Na₃PO₄, clearly AgNO₃ is limiting as it will run out first.

a) 0.002505 mol AgNO₃ x 1 mol Ag₃PO₄ / 3 mol AgNO₃ x 419 g Ag₃PO₃ / mol = 0.350 g Ag₃PO₄ obtained

b) Excess substance is Na₃PO₄. Concentration remaining after reaction is as follows: Find moles Na₃PO₄ used and subtract that from moles Na₃PO₄ initially present and then convert to molarity by dividing by liters.

0.002505 mol AgNO₃ x 1 mol Na₃PO₄ used / 3 mol AgNO₃ = 0.000835 mol Na₃PO₄ used

0.01498 mol - 0.000835 mol = 0.01415 moles Na₃PO₄ remaining

Final volume = 35.0 ml + 15.0 ml = 50.0 ml = 0.05 L

[Na₃PO₄] after precipitation = 0.01415 mol / 0.05 L = 0.283 M = [Na₃PO₄]