Question

How many grams of barium sulfate, BaSO 4 , be precipitated when 2.25 moles of sodium sulfate , Na 2 SO 3 , reacts with an excess of barium chloride in solution ?

Answer 1

Answer:

525.1 g of BaSO₄ are produced.

Explanation:

The reaction of precipitation is:

Na₂SO₄ (aq) + BaCl₂ (aq) →  BaSO₄ (s) ↓  +  2NaCl (aq)

Ratio is 1:1. So 1 mol of sodium sulfate can make precipitate 1 mol of barium sulfate.

The excersise determines that the excess is the  BaCl₂.

After the reaction goes complete and, at 100 % yield reaction, 2.25 moles of BaSO₄ are produced.

We convert the moles to mass: 2.25 mol . 233.38 g/mol = 525.1 g

The precipitation's equilibrium is:

SO₄⁻² (aq)  +  Ba²⁺ (aq)  ⇄  BaSO₄ (s) ↓       Kps