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Answer :
Final answer:
Calculating the grams of H₂SO₄ needed to react with 89.1 grams of Al requires stoichiometry and gives a result of approximately 486.0 grams. None of the given options are correct.
Explanation:
To determine how many grams of sulfuric acid (H₂SO₄) are needed to completely react with 89.1 grams of aluminum (Al) in the reaction we need to use stoichiometry.
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂
Calculate the Molar Mass of Aluminum (Al):
Aluminum (Al) has a molar mass of approximately 26.98 g/mol.
Convert the Mass of Al to Moles:
Convert the mass of Al (89.1 g) to moles by dividing by its molar mass:
Moles of Al = Mass of Al/ Molar mass of Al
Moles of Al = 89.1 g / 26.98 g/mol = 3.303 moles of Al.
Use the Stoichiometry of the Reaction:
The balanced chemical equation shows that 2 moles of Al react with 3 moles of H₂SO₄.
Therefore, the amount of H₂SO₄ needed for 3.303 moles of Al can be calculated by:
Moles of H₂SO₄ = 3.303 moles of Al x [tex]\frac{3 moles of H_2SO_4}{2 moles of Al}[/tex] = 4.9545 moles of H₂SO₄
Convert Moles of H₂SO₄ to Mass:
The molar mass of sulfuric acid (H₂SO₄) is approximately 98.079 g/mol.
To find the mass of H₂SO₄ required:
Mass of H₂SO₄ = 4.9545 moles of H₂SO₄ x 98.079 g/mol ≈ 486.0 grams
Therefore, approximately 486.0 grams of H₂SO₄ are required to completely react with 89.1 grams of aluminum.
None of the given options are correct.
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