Question

400. mL of 0.5875 M aqueous silver nitrate are mixed with 500. mL of 0.290 M sodium chromate.

a) What is the mass of the precipitate?

b) What is the concentration of each of the ions remaining in solution? Indicate any assumptions or approximations.

Answer 1

(a)

2 AgNO3 (aq.) + Na2CrO4 (aq.) -----------> Ag2CrO4 (s) + 2 NaNO3 (aq.)

Moles of AgNO3 = 0.5875 * 400. / 1000 = 0.235 mol

Moles of sodium chromate = 0.290 * 500 / 1000 = 0.145 mol

From the balanced equation,

2 mol of agNO3 needs 1 mol Na2CrO4

Then,

0.235 mol of AgNO3 needs 1 * 0.235 / 2 = 0.1175 mol of Na2CrO4

But we have 0.145 mol of Na2CrO4 , it is excess reagent.

so, AgNO3 is limiting reagent.

From the balanced equation,

2 mol AgNO3 forms 1 mol of precipitate

2*169.87 g. of AgNO3 forms 331.73 g.

Then,

0.235 mol forms * 331.73 * 0.235 / 2 = 39.0 g. of precipitate.

(b)

Remaining concentration of

Na2CrO4 = 0.145 - 0.1175 = 0.0275 mol

[Na+] = 2 * 0.0275 * 1000 / (400. + 500.) = 0.0611 M

[CrO4] = 0.0275 * 1000 / 900 = 0.0306 M

[NO3-] = 0.235 * 1000 / 900 = 0.261 M