Final answer:
To produce 4.00 L of hydrogen gas at STP, 3.21 grams of aluminum are required based on the reaction between aluminum and sulfuric acid and using stoichiometry calculations.
Step-by-step explanation:
To determine the mass of aluminum needed to produce 4.00 L of hydrogen gas at STP (Standard Temperature and Pressure), we use the stoichiometry of the given reaction:
2 Al(s) + 3H₂SO₄ (aq) → Al₂(SO₄)₃ (aq) + 3H₂(g)
From Avogadro's law, we know that one mole of a gas occupies 22.4 L at STP. Therefore, the moles of hydrogen gas produced can be calculated by dividing the volume of hydrogen gas at STP by the molar volume:
Number of moles of H₂ = 4.00 L / 22.4 L/mol = 0.17857 mol H₂
Using the stoichiometry of the reaction, we find that 3 moles of H₂ is produced from 2 moles of Al. Thus, for every 1.5 moles of H₂, 1 mole of Al is required. We can use this ratio to calculate the moles of Al needed to produce the hydrogen:
Moles of Al needed = (0.17857 mol H₂) * (2 mol Al / 3 mol H₂) = 0.11905 mol Al
To find the mass of aluminum, we multiply the moles of Al by its molar mass (26.98 g/mol):
Mass of Al = 0.11905 mol Al * 26.98 g/mol = 3.21 g
Therefore, 3.21 grams of aluminum are needed to produce 4.00 L of hydrogen gas at STP.