Final answer:
To react with 28.5 grams of aluminum, at least 1140 mL of 1.39 M copper(II) sulfate solution is required.
Step-by-step explanation:
To determine the minimum volume of copper(II) sulfate solution required to react with 28.5 grams of aluminum metal, we need to use stoichiometry. The balanced equation for the reaction is:
2 Al(s) + 3 CuSO₄(aq) → Al2(SO₄)₃(aq) + 3 Cu(s)
From the equation, we can see that 3 moles of copper(II) sulfate react with 2 moles of aluminum. To calculate the moles of aluminum, we divide the given mass (28.5 grams) by the molar mass of aluminum:
(28.5 g Al) / (26.98 g/mol Al) = 1.058 mol Al
Since the mole ratio between aluminum and copper(II) sulfate is 2:3, the moles of copper(II) sulfate needed is:
(1.058 mol Al) × (3 mol CuSO₄/ 2 mol Al) = 1.587 mol CuSO₄
To convert moles of copper(II) sulfate to volume, we need to use the concentration of the solution. The given concentration is 1.39 M, which means there are 1.39 moles of copper(II) sulfate in 1 liter of solution.
So, the minimum volume of 1.39 M copper(II) sulfate required is:
(1.587 mol CuSO₄) / (1.39 mol/L) = 1.14 L = 1140 mL