A bracelet, originally made of pure silver, became tarnished over time with black silver sulfide (Ag₂S) forming on the surface - Leaving Cert Chemistry - Question c - 2012

To find the moles of sulfur removed, we first calculate the mass decrease of the bracelet:

$ext{Mass decrease} = 0.006\% \times \text{mass of bracelet}$ Assuming the initial mass of the bracelet is 100 g (for simplicity), the mass decrease is:

$0.006\% \times 100 g = 0.006 g$

Now, since each mole of Ag₂S contains 1 mole of S and considering the molar mass of S as 32 g/mol:

$\text{Moles of S} = \frac{0.006 g}{32 g/mol} = 0.0003 mol$

According to the balanced reaction, 2 moles of Al are required for every 3 moles of Ag₂S. Thus, for 0.0003 moles of sulfur, the moles of aluminum can be calculated as follows:

$\frac{0.0003\, mol\, S}{0.003\, mol\, Ag_2S} \times \frac{2\, mol\, Al}{3\, mol\, Ag_2S} = 0.0002 mol\, Al$

The molar mass of aluminum is approximately 27 g/mol, so:

$\text{Mass of Al} = 0.0002 mol \times 27 g/mol = 0.0054 g$

To find the loss in mass if the entire amount of silver sulfide was removed, we use the moles of sulfur:

$\text{Moles of S} = 0.0003 mol\, S$

Using the stoichiometry of Ag₂S, we know that:

$\text{Moles of Ag}_2S = 0.0003 mol\, S \times \frac{1 mol\, Ag_2S}{1 mol\, S} = 0.0003 mol\, Ag_2S$

We then calculate the total mass of the silver sulfide:

$0.0003 mol\, Ag_2S \times 248 g/mol \approx 0.0744 g$

Thus, the loss in mass would be approximately 0.0744 g.