Answer: The balanced chemical equation for the reaction between aluminum (Al) and copper(II) sulfate (CuSO4) is:
2 Al (s) + 3 CuSO4 (aq) → Al2(SO4)3 (aq) + 3 Cu (s)
From the balanced equation, we can see that 2 moles of Al react with 3 moles of CuSO4 to produce 1 mole of Al2(SO4)3 and 3 moles of Cu. We can use this information to calculate the theoretical yield of Al2(SO4)3 that can be produced from a given amount of Al.
To do this, we first need to convert the mass of Al given in the problem to moles using its molar mass. The molar mass of Al is 26.98 g/mol, so:
459.7 g Al × (1 mol Al / 26.98 g Al) = 17.03 mol Al
We can then use the mole ratio from the balanced equation to calculate the moles of Al2(SO4)3 that can be produced:
2 mol Al : 1 mol Al2(SO4)3
17.03 mol Al : x mol Al2(SO4)3
x = (17.03 mol Al) / 2 × (1 mol Al2(SO4)3 / 1 mol Al) = 8.52 mol Al2(SO4)3
Finally, we can convert the moles of Al2(SO4)3 to grams using its molar mass:
Molar mass of Al2(SO4)3 = 342.15 g/mol
Mass of Al2(SO4)3 = 8.52 mol Al2(SO4)3 × (342.15 g/mol) = 2915 g
Therefore, approximately 2915 grams of aluminum sulfate (Al2(SO4)3) are produced from 459.7 grams of aluminum (Al) reacting with copper(II) sulfate (CuSO4).