Answer: 37.1 grams of aluminum hydroxide can be produced from the given amounts of aluminum oxide and water
Step-by-step explanation:
The balanced chemical equation for the synthesis of aluminum hydroxide from aluminum oxide and water is:
Al2O3 + 3H2O -> 2Al(OH)3
From the equation, we can see that 1 mole of Al2O3 reacts with 3 moles of H2O to produce 2 moles of Al(OH)3.
We need to first calculate the number of moles of Al2O3 and H2O we have:
Number of moles of Al2O3 = mass / molar mass = 24.3 g / 101.96 g/mol = 0.238 moles
Number of moles of H2O = mass / molar mass = 82.1 g / 18.02 g/mol = 4.56 moles
From the balanced equation, we know that 1 mole of Al2O3 produces 2 moles of Al(OH)3. So, the number of moles of Al(OH)3 produced will be:
Number of moles of Al(OH)3 = (0.238 mol Al2O3) x (2 mol Al(OH)3 / 1 mol Al2O3) = 0.476 mol Al(OH)3
Finally, we can calculate the mass of Al(OH)3 produced:
Mass of Al(OH)3 = number of moles x molar mass = 0.476 mol x 78.0 g/mol = 37.1 g
Therefore, 37.1 grams of aluminum hydroxide can be produced from the given amounts of aluminum oxide and water.