Calculate the mass of the precipitate that forms when an excess of sodium phosphate solution

Calculate the mass of the precipitate that forms when an excess of sodium phosphate solution is added to 17.1 g of a 30% silver nitrate solution.

Given:

m (AgNO3) = 17.1 g

w% (AgNO3) = 30%

To find:

m (Ag3PO4) -?

Decision:

3AgNO3 + Na3PO4 = Ag3PO4 + 3NaNO3, – we solve the problem, relying on the composed reaction equation:

1) Find the mass of silver nitrate that has reacted:

m (AgNO3) = 17.1 g * 0.3 = 5.13 g

2) Find the amount of silver nitrate:

n (AgNO3) = m: M = 5.13 g: 170 g / mol = 0.03 mol.

3) We compose a logical expression:

If 3 mol of AgNO3 gives 1 mol of Ag3PO4,

then 0.03 mol of AgNO3 gives x mol of Ag3PO4,

then x = 0.01 mol.

4) Find the mass of precipitated silver phosphate:

m (Ag3PO4) = n * M = 0.01 mol * 419 g / mol = 4.19 g.

Answer: m (Ag3PO4) = 4.19 g.