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Answer :
The limiting reagent for the reaction between 513 grams of barium hydroxide and 98.1 grams of sulfuric acid is sulfuric acid (H₂SO₄).
1) To identify the limiting reagent in the given reaction, we must first determine the moles of each reactant.
Starting with barium hydroxide (Ba(OH)₂):
- Molar mass of Ba(OH)₂ = 137.33 (Ba) + 2 × 17.00 (OH)
- = 171.33 g/mol
- Moles of Ba(OH)₂ = 513 g / 171.33 g/mol
- ≈ 2.99 moles
2) Next, for sulfuric acid (H₂SO₄):
- Molar mass of H₂SO₄ = 2 × 1.01 (H) + 32.07 (S) + 4 × 16.00 (O)
- = 98.09 g/mol
- Moles of H₂SO₄ = 98.1 g / 98.09 g/mol
- ≈ 1.00 moles
The balanced equation is:
Ba(OH)₂ (aq) + H₂SO₄ (aq) → BaSO₄ (s) + 2 H₂O (l)
From the balanced equation, the mole ratio of Ba(OH)₂ to H₂SO₄ is 1:1. To react with 2.99 moles of Ba(OH)₂, we would need 2.99 moles of H₂SO₄. The limiting reagent is is sulfuric acid (H₂SO₄).
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