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.

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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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