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
525.1 0 g of BaSO₄ are produced.
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 of 1 mol of barium sulfate.
Molar mass of [tex]BaSO_4[/tex] is 233.38 g/mol
The formula for the number of moles is as follows:-
[tex]Number\ of\ moles=\frac{Mass}{Molar\ Mass}[/tex]
[tex]Mass=2.25\ moles\times233.38\ g/mol\\=525.10\ g[/tex]
So, 525.10 g of barium sulfate be precipitated.
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