If your favorite silver jewelry ever tarnishes and you happen to have some aluminum foil on hand, you can actually get rid of some of that tarnish. The tarnish here is actually silver sulfide. If you put the jewelry and aluminum into a dish with enough water to cover them both, the following reaction will occur: 3Ag2S + 2Al --> 6Ag + Al2S3. If you have a total of 5.0 g of Al foil accessible to you, how many grams of Ag2S (thus, loosely speaking, how much jewelry) could you theoretically "clean"?
how many moles in 5g of Al?
The equation says you will use 3/2 that many moles of Ag2S
So, convert that back to grams of Ag2S
of course, the amount of jewelry will be proportional to its surface area, not its volume.
To determine how many grams of Ag2S you could theoretically clean using 5.0 g of aluminum (Al) foil, we need to use stoichiometry, which involves converting between the moles of the given substances.
From the balanced chemical equation provided:
3Ag2S + 2Al --> 6Ag + Al2S3
We can see that the molar ratio between Ag2S and Al is 3:2.
This means that for every 3 moles of Ag2S, we would need 2 moles of Al.
First, we will calculate the number of moles of Al available.
Given: Mass of Al foil = 5.0 g
Atomic mass of Al = 26.98 g/mol
Number of moles of Al = Mass of Al foil / Atomic mass of Al
= 5.0 g / 26.98 g/mol
≈ 0.185 mol
Now, using the stoichiometric ratio, we can find the number of moles of Ag2S that can be cleaned:
Number of moles of Ag2S = (Number of moles of Al / 2) * (3 moles of Ag2S / 2 moles of Al)
≈ 0.185 mol / 2 * 3/2
≈ 0.278 mol
To convert the number of moles of Ag2S into grams, we will use the molar mass of Ag2S.
Molar mass of Ag2S = (2 x atomic mass of Ag) + atomic mass of S
= (2 x 107.87 g/mol) + 32.07 g/mol
≈ 242.81 g/mol
Mass of Ag2S = Number of moles of Ag2S x Molar mass of Ag2S
= 0.278 mol x 242.81 g/mol
≈ 67.42 g
Therefore, with 5.0 g of aluminum foil, you could theoretically clean approximately 67.42 grams of Ag2S or tarnished jewelry.