aluminum reacts with sulfuric acid to produce aluminum sulfate and hydrogen gas. how many grams of aluminum sulfate would be formed if 250g H2SO4 reacted with aluminum?

Max, I worked a similar problem, the one with Fe and Fe2O3, for you when you were Campbell. Just follow the same procedure.

spelled Cambell.

To determine the amount of aluminum sulfate formed when 250g of sulfuric acid (H2SO4) reacts with aluminum, we need to follow a few steps:

Step 1: Write the balanced chemical equation:
2Al + 3H2SO4 → Al2(SO4)3 + 3H2

The equation tells us that 2 moles of aluminum (Al) react with 3 moles of sulfuric acid (H2SO4) to produce 1 mole of aluminum sulfate (Al2(SO4)3) and 3 moles of hydrogen gas (H2).

Step 2: Convert the mass of H2SO4 (in grams) to moles.
To do this, we need to know the molecular weight (molar mass) of H2SO4, which is 98.09 g/mol.

Number of moles of H2SO4 = mass / molar mass = 250g / 98.09 g/mol ≈ 2.55 moles

Step 3: Determine the stoichiometric ratio between H2SO4 and Al2(SO4)3.
From the balanced equation, we can see that 3 moles of H2SO4 react to produce 1 mole of Al2(SO4)3.

Step 4: Calculate the number of moles of Al2(SO4)3 formed.
For every 3 moles of H2SO4, we have 1 mole of Al2(SO4)3.
So, moles of Al2(SO4)3 = (moles of H2SO4) / (3 moles of H2SO4 / 1 mole of Al2(SO4)3) = 2.55 moles / (3/1) ≈ 0.85 moles

Step 5: Convert moles of Al2(SO4)3 to grams.
To do this, we need to know the molar mass of Al2(SO4)3, which is approximately 342.15 g/mol.

Mass of Al2(SO4)3 = moles of Al2(SO4)3 * molar mass = 0.85 moles * 342.15 g/mol ≈ 291.33 g

Therefore, approximately 291.33 grams of aluminum sulfate would be formed when 250 grams of sulfuric acid reacts with aluminum.