To solve this problem, we will first write a balanced chemical equation for the reaction between aluminum (Al) and hydrochloric acid (HCl) in the presence of water (H2O). The reaction produces aluminum chloride (AlCl3) and hydrogen gas (H2). The balanced equation is:
2Al (s) + 6HCl (aq) → 2AlCl3 (aq) + 3H2 (g)
Next, we will use stoichiometry to find the amount of hydrogen gas (H2) produced. First, we need to convert the mass of aluminum (13.55 g) into moles. To do this, we will use the molar mass of aluminum, which is approximately 26.98 g/mol.
moles of Al = mass of Al / molar mass of Al
moles of Al = 13.55 g / 26.98 g/mol
moles of Al ≈ 0.501 moles
Now, we can use the stoichiometry from the balanced equation to find the moles of H2 produced:
2 moles of Al react with 3 moles of H2, so:
moles of H2 = (moles of Al × moles of H2) / moles of Al
moles of H2 = (0.501 moles × 3) / 2
moles of H2 ≈ 0.7515 moles
Finally, we will convert the moles of H2 into grams using the molar mass of hydrogen gas, which is approximately 2.016 g/mol:
mass of H2 = moles of H2 × molar mass of H2
mass of H2 ≈ 0.7515 moles × 2.016 g/mol
mass of H2 ≈ 1.514 g
So, approximately 1.514 grams of hydrogen gas (H2) will be produced when 13.55 grams of aluminum reacts with hydrochloric acid under the given conditions.
