Question

Silver tarnishes in the presence of hydrogen sulfide (which smells like rotten eggs) and oxygen because of the reaction:

4Ag + 2H₂S + O₂ → 2Ag₂S + 2H₂O

a) How many grams of silver sulfide can be formed from a mixture of 1.1 g Ag, 0.10 g H₂S, and 0.088 g O₂?

b)How many more grams of H₂S would be needed to completely react all of the Ag?

Answer 1

a) Given reaction

4Ag + 2H2S + O2 = 2Ag2S + 2H2O

The amount of substances is

n(mol) = m(g) / MW(g/mol)

n(Ag) = m(Ag) / MW(Ag) = 1.1 / 108 = 0.010mol

n(H2S) = m(H2S) / MW(H2S) = 0.1 / (34) =0.00294 mol

n(O2) = m(O2) / MW(O2) = 0.088 / 32 = 0.00275 mol

According to the equation

n(Ag) : n(H2S) : n(O2) = 4: 2 : 1

But actually we have n(Ag): n(H2S) : n(O2) = 4 : 1.17 : 1.1

As we can see the amount of H2S is not enough, so H2S is limiting reagent.

n(Ag2S) = n(H2S) = 0.00294 mol (from the reaction)

The weight of Ag2S is

m(Ag2S) = n(Ag2S) x MW(Ag2S) = 0.00294 x (248) = 0.729 g

m(Ag2S) = 0.724 g of silver sulfide is formed.

b) n(Ag) = m(Ag) / MW(Ag) = 1.1 / 108 = 0.010mol

n(H2S) = m(H2S) / MW(H2S) = 0.1 / (34) =0.00294 mol

mole ratio between Ag and H2S is 4:2

That means 0.04 moles of Ag needs 0.02moles of H2S, but here it has 0.0294moles.

0.68 g of H2S required, but it has only 0.1 g, so 0.58 grams of H2S would be needed to completely react all of the Ag