Final answer:
Using stoichiometry, we determine that 216 grams of Al are needed to react with excess HCl to produce 24.0 grams of H2 based on the balanced equation provided.
Step-by-step explanation:
To determine how many grams of Al are needed to react with excess HCl to produce 24.0 grams of H2, we can use stoichiometry based on the balanced chemical equation 2Al + 6 HCl → 2 AlCl3 + 3 H2.
First, we calculate the number of moles of H2 produced using the molar mass of H2 (approximately 2 g/mol). Then, using the stoichiometry of the balanced equation, we find the mole ratio between H2 and Al, which is 3:2.
This means that for every 3 moles of H2, 2 moles of Al are consumed.
Since 24.0 grams of H2 is 24.0 g / 2 g/mol = 12.0 moles of H2,
the number of moles of Al required is (2/3) × 12.0 moles = 8.0 moles of Al.
Then, we calculate the mass of Al needed using its molar mass (approximately 27 g/mol),
which gives us 8.0 moles × 27 g/mol = 216 grams of Al.