Final answer:
The mass of aluminum required to produce 560.0 mL of hydrogen gas at STP is 8.0 g.
Step-by-step explanation:
To determine the mass of aluminum required to produce 560.0 mL of hydrogen gas at STP, we need to use the stoichiometry of the given balanced equation: 2Al(s) + 6HCl(aq) ⟶ 2AlCl₃(aq) + 3H₂(g).
From the balanced equation, we can see that 3 moles of hydrogen gas are produced for every 2 moles of aluminum. We can use this ratio to find the moles of hydrogen gas produced using the ideal gas law.
Since the volume of hydrogen gas is given in mL, we need to convert it to liters by dividing by 1000. Then, we can use the equation PV = nRT to find the moles of hydrogen gas. Once we have the moles of hydrogen gas, we can use the mole ratio to find the moles of aluminum. Finally, we can convert the moles of aluminum to grams using its molar mass of 26.98 g/mol.
Therefore, the mass of aluminum required to produce 560.0 mL of H₂(g) at STP is approximately 8.0 g.