To 240 g of a 16.4% sodium phosphate solution was added 280 g of a 16% solution of calcium bromide.

To 240 g of a 16.4% sodium phosphate solution was added 280 g of a 16% solution of calcium bromide. Calculate the mass of the precipitate formed.

Given:

m (Na3PO4) = 240 g

m (CaBr2) = 280 g

w% (Na3PO4) = 16.4%

w% (CaBr2) = 16%

To find:

m (draft) -?

Decision:

3CaBr2 + 2Na3PO4 = Ca3 (PO4) 2 + 6NaBr, – we solve the problem, relying on the composed reaction equation:

1) Find the masses of sodium phosphate and calcium bromide:

m (Na3PO4) = 240 g * 0.164 = 39.36 g

m (CaBr2) = 280 g * 0.16 = 44.8 g

2) Find the amount of sodium phosphate and calcium bromide:

n (Na3PO4) = m: M = 39.36 g: 164 g / mol = 0.24 mol

n (CaBr2) = m: M = 44.8 g: 200 g / mol = 0.224 mol

We start from a lower value to get more accurate calculations. We work with CaBr2:

3) We compose a logical expression:

if 3 mol CaBr2 gives 1 mol Ca3 (PO4) 2,

then 0.224 mol CaBr2 will give x mol Ca3 (PO4) 2,

then x = 0.075 mol.

4) Find the mass of calcium phosphate precipitated during the reaction:

m (Ca3 (PO4) 2) = n * M = 0.075 mol * 310 g / mol = 23.25 g.

Answer: m (Ca3 (PO4) 2) = 23.25 g.