Calculate the mass of aluminum sulfate that will be formed if 2.531 grams of aluminum hydroxide reacts with 3.892 grams of sulfuric acid. The chemical reaction (unbalanced) is
Al(OH)3 + H2SO4 → Al2(SO4)3 + H2O
i know that the balanced formula is 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 +6H2O
need help with chemistry problem?
2017-08-24 3:15 pm
回答 (3)
2017-08-24 4:24 pm
According to Wikipedia :
Molar mass of Al(OH)₃ = 78.00 g/mol
Molar mass of H₂SO₄ = 98.08 g/mol
Molar mass of Al₂(SO₄)₃ = 342.15 g/mol
Initial number of moles of Al(OH)₃ = (2.531 g) / (78.00 g/mol) = 0.03245 mol
Initial number of moles of H₂SO₄ = (3.892 g) / (98.08 g/mol) = 0.03968 mol
Balanced equation for the reaction :
2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O
Mole ratio Al(OH)₃ : H₂SO₄ : Al₂(SO₄)₃ = 2 : 3 : 1
When 0.03968 mol H₂SO₄ completely reacts :
Number of moles of Al(OH)₃ needed = (0.03968 mol) × (2/3) = 0.02645 mol < 0.03245 mol
Al(OH)₃ is in excess. Then, H₂SO₄ completely reacts (the limiting reactant).
In the reaction :
Number of moles of H₂SO₄ reacted = 0.03968 mol
Number of moles of Al₂(SO₄)₃ formed = (0.03968 mol) × (1/3) = 0.01323 mol
Mass of Al₂(SO₄)₃ formed = (0.01323 mol) × (342.15 g/mol) = 4.527 g
Molar mass of Al(OH)₃ = 78.00 g/mol
Molar mass of H₂SO₄ = 98.08 g/mol
Molar mass of Al₂(SO₄)₃ = 342.15 g/mol
Initial number of moles of Al(OH)₃ = (2.531 g) / (78.00 g/mol) = 0.03245 mol
Initial number of moles of H₂SO₄ = (3.892 g) / (98.08 g/mol) = 0.03968 mol
Balanced equation for the reaction :
2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O
Mole ratio Al(OH)₃ : H₂SO₄ : Al₂(SO₄)₃ = 2 : 3 : 1
When 0.03968 mol H₂SO₄ completely reacts :
Number of moles of Al(OH)₃ needed = (0.03968 mol) × (2/3) = 0.02645 mol < 0.03245 mol
Al(OH)₃ is in excess. Then, H₂SO₄ completely reacts (the limiting reactant).
In the reaction :
Number of moles of H₂SO₄ reacted = 0.03968 mol
Number of moles of Al₂(SO₄)₃ formed = (0.03968 mol) × (1/3) = 0.01323 mol
Mass of Al₂(SO₄)₃ formed = (0.01323 mol) × (342.15 g/mol) = 4.527 g
2017-08-24 3:58 pm
2.531 g Al(OH)₃ x (1 mol Al(OH)₃ / 77.98 g Al(OH)₃) x (3 mol H₂SO₄ / 2 mol Al(OH)₃) x (98 g H₂SO₄ / 1 mol H₂SO₄) = 4.771 g H₂SO₄ required --> Al(OH)₃ is in excess
3.892 g H₂SO₄ x (1 mol H₂SO₄ / 98 g H₂SO₄) x (1 mol Al₂(SO₄)₃) / 3 mol H₂SO₄) x (342.16 g Al₂(SO₄)₃ / 1 mol Al₂(SO₄)₃) = 4.529 g Al₂(SO₄)₃ required
3.892 g H₂SO₄ x (1 mol H₂SO₄ / 98 g H₂SO₄) x (1 mol Al₂(SO₄)₃) / 3 mol H₂SO₄) x (342.16 g Al₂(SO₄)₃ / 1 mol Al₂(SO₄)₃) = 4.529 g Al₂(SO₄)₃ required
2017-08-24 3:16 pm
you should contact my chemistry professor
收錄日期: 2021-05-01 13:08:48
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