Final answer:
To react completely with 107 mL of 6.00 M H₂SO₄, we must convert the volume of acid to moles and then use stoichiometry based on the balanced equation. The calculation reveals that 0.428 moles of aluminum are required.
Step-by-step explanation:
To determine the number of moles of aluminum required to react with the given amount of sulfuric acid, we will use the stoichiometry of the balanced chemical equation.
From the balanced equation: 2 Al(s) + 3 H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3 H₂(g)
It is given that the sulfuric acid has a concentration of 6.00 M.
This means that there are 6.00 moles of sulfuric acid in 1 liter of the solution.
First, we need to convert the volume of sulfuric acid from mL to L:
107 mL × (1 L / 1000 mL) = 0.107 L
Next, we use the concentration of sulfuric acid to calculate the moles:
6.00 M × 0.107 L = 0.642 moles of H₂SO₄
Finally, we use the stoichiometry of the balanced equation to determine the moles of aluminum:
3 moles of H₂SO₄ react with 2 moles of Al
0.642 moles of H₂SO₄ × (2 moles of Al / 3 moles of H₂SO₄) = 0.428 moles of Al
Therefore, 0.428 moles of aluminum are required to completely react with 107 mL of 6.00 M H₂SO₄.