# Aluminum reacts with hydrochloric acid to produce aluminum chloride and hydrogen gas. How many liters of 6.0 M hydrochloric acid would you need to completely react with the roll of aluminum foil? The roll contains 23.226 square meters of aluminum foil.

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.

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