Final answer:
The calculated mass of Al₂(SO₄)₃ produced from reacting 152 g of H₂SO₄ is 176.69 g. However, this value is not listed in the multiple-choice options provided, suggesting an error in the question or answer options.
Step-by-step explanation:
In the reaction 2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O, to calculate the mass of Al₂(SO₄)₃ produced from 152 g of H₂SO₄, we'll first convert the mass of H₂SO₄ to moles using its molar mass. The molar mass of H₂SO₄ is 98.079 g/mol, so:
152 g H₂SO₄ x (1 mol H₂SO₄ / 98.079 g H₂SO₄) = 1.549 moles H₂SO₄
Next, we use the stoichiometry of the reaction which indicates that 3 moles of H₂SO₄ produce 1 mole of Al₂(SO₄)₃. Therefore, the moles of Al₂(SO₄)₃ produced are:
1.549 moles H₂SO₄ x (1 mol Al₂(SO₄)₃ / 3 moles H₂SO₄) = 0.5163 moles Al₂(SO₄)₃
The molar mass of Al₂(SO₄)₃ is 342.15 g/mol, so the mass of Al₂(SO₄)₃ produced is:
0.5163 moles Al₂(SO₄)₃ x (342.15 g Al₂(SO₄)₃ / 1 mole Al₂(SO₄)₃) = 176.69 g Al₂(SO₄)₃
Since this result is not an option in the provided choices, and assuming this is a real situation or an error within the question, the calculation should be re-examined for potential mistakes. However, based on the information given and correct calculations, none of the options provided (A, B, C, D) accurately reflect the calculated mass of aluminum sulfate.