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

See example below.

To determine the grams of aluminum sulfate formed, we need to follow a step-by-step process:

Step 1: Write and balance the chemical equation:
The balanced equation for the reaction between aluminum and sulfuric acid is:
2 Al + 3 H2SO4 → Al2(SO4)3 + 3 H2

Step 2: Determine the molar mass:
The molar mass of sulfuric acid (H2SO4) is:
Molar mass of H2SO4 = 2(1.01 g/mol for hydrogen) + 32.07 g/mol for sulfur + 4(16.00 g/mol for oxygen) = 98.09 g/mol

Step 3: Convert grams to moles of H2SO4:
To convert grams to moles, divide the given mass by the molar mass:
Moles of H2SO4 = 250 g H2SO4 / 98.09 g/mol ≈ 2.55 moles

Step 4: Determine the stoichiometric ratio:
From the balanced equation, we can see that 1 mole of H2SO4 reacts with 1 mole of Al2(SO4)3. Therefore, the stoichiometric ratio of H2SO4 to Al2(SO4)3 is 1:1.

Step 5: Convert moles of H2SO4 to moles of Al2(SO4)3:
Since the stoichiometric ratio is 1:1, the number of moles of H2SO4 is the same as the moles of Al2(SO4)3.

Step 6: Determine the molar mass of Al2(SO4)3:
The molar mass of aluminum sulfate (Al2(SO4)3) is:
2(26.98 g/mol for aluminum) + 3(32.07 g/mol for sulfur) + 12(16.00 g/mol for oxygen) = 342.14 g/mol

Step 7: Convert moles of Al2(SO4)3 to grams:
To convert moles to grams, multiply the moles by the molar mass:
Grams of Al2(SO4)3 = 2.55 moles Al2(SO4)3 × 342.14 g/mol ≈ 871.94 g

Therefore, approximately 871.94 grams of aluminum sulfate would be formed if 250 grams of H2SO4 completely reacted with aluminum.