Final answer:
The mass of sulfuric acid needed to react with 89.1g of aluminum can be found by using stoichiometry and the balanced equation 2 Al + 3 H₂SO₄ → Al₂(SO₄)₃ + 3 H₂. By calculating moles of aluminum and converting it to moles of sulfuric acid and then to mass, we can find the required mass of sulfuric acid.
Step-by-step explanation:
To determine the mass of sulfuric acid needed to completely react with 89.1g of aluminum, we first need to write a balanced chemical equation for the reaction:
2 Al(s) + 3 H₂SO₄ (aq) → Al₂(SO₄)₃ (aq) + 3 H₂(g)
From the equation, 2 moles of aluminum (Al) react with 3 moles of sulfuric acid (H₂SO₄). First, we calculate the molar mass of aluminum, which is approximately 26.98 g/mol, and that of sulfuric acid, which is approximately 98.08 g/mol.
The molar mass of Al, which is 26.98g/mol, suggests that 2 moles of Al weigh 53.96g. The equation shows that 2 moles of Al react with 3 moles of H₂SO₄; so for 89.1g of Al, we use stoichiometry to find the moles of H₂SO₄:
89.1g Al * (1 mol Al/26.98g Al) * (3 mol H₂SO₄/2 mol Al) = x mol H₂SO₄
Now multiply by the molar mass of H₂SO₄ to find the mass:
x mol H₂SO₄ * 98.08g H₂SO₄/mol = grams of sulfuric acid
After completing the calculations, the mass of sulfuric acid required will be determined, and our answer would be expressed to three significant figures.