This question is based on stoichiometry, and hence we need a balanced chemical equation for the reaction. The reaction can be written as the following:
2Al + 6HCl -> 2AlCl3 + 3H2
Here, 2 moles of aluminum reacts with 6 moles of hydrochloric acid to generate 3 moles of aluminum chloride and 3 moles of hydrogen gas. In other words, 1 mole of aluminum reacts with 3 moles of hydrochloric acid.
Let's determine the moles of aluminum in the given length of foil. The thickness of a standard household aluminum foil is typically 0.016 mm (or 0.016 x 10^-3 m).
The given area of the foil roll is 23.226 m2.
The volume of the foil roll = area x thickness = 23.226 m2 x 0.016 x 10^-3 m
= 3.72 x 10^-4 m3 = 372 cm3
(since 1 m3 = 10^6 cm3).
Now the density of aluminum is about 2.7 g/cm3.
Since, density = mass/volume; mass = density x volume.
Thus, the mass of aluminum foil roll = 2.7 g/cm3 x 372 cm3 = 1004.4 g.
The atomic mass of aluminum is 27 g/mole.
Thus, the moles of aluminum in the foil roll = 1004.4/27 = 37.2 moles.
Since each mole of aluminum reacts with 3 moles of HCl, the moles of HCl that will react with 37.2 moles of aluminum are equal to 111.6 moles (= 37.2 x 3).
Molarity = moles / volume
or, Volume = moles/molarity = 111.6 moles /6 M = 18.6 l.
Thus, one will need 18.6 liters of 6M HCl to completely react with a roll of aluminum foil of standard thickness.
Hope this helps.