Final answer:
To neutralize 96.5 mL of 0.556 M H2SO4, 1.395 grams of Al(OH)3 are required. The calculation is based on the stoichiometry of the reaction between Al(OH)3 and H2SO4.
Step-by-step explanation:
To calculate the mass of Al(OH)3 needed to neutralize a given volume of 0.556 M H2SO4, we must first write the balanced equation for the reaction:
Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
From the balanced equation, we see that 1 mole of Al(OH)3 reacts with 3 moles of H2SO4. Using the molarity and volume of H2SO4, we calculate the moles of H2SO4:
0.556 moles/L * 0.0965 L = 0.053654 moles of H2SO4
Then, we use the stoichiometric relationships to find the moles of Al(OH)3 needed:
0.053654 moles H2SO4 * (1 mole Al(OH)3 / 3 moles H2SO4) = 0.01788467 moles of Al(OH)3
The molar mass of Al(OH)3 is approximately 78.00 g/mol, so the mass needed is:
0.01788467 moles * 78.00 g/mol = 1.395 g of Al(OH)3