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

Question

2 years ago

15

of barium chloride in solution ?

1 answer:

gulaghasi [49] · Answer 1 · 2022-11-29T23:34:10+03:00

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