Answer to Question #25715 in Inorganic Chemistry for Chem

Question

Silver tarnishes in the presence of hydrogen sulfide (which smells like rotten eggs) and oxygen because of the following reaction: Ag + H2S + O2 --> Ag2S + H2O .

a) How many grams of silver sulfide can be formed from a mixture of 1.1 g Ag, 0.14 g H2S, and 0.080 g O2?



b) How many more grams of hydrogen sulfide would be needed to completely react all the silver?

Answer 1

4Ag + 2H₂S + O₂ = 2Ag₂S + 2H₂O

The amount of substances is

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

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

n(H₂S) = m(H₂S) / M_W(H₂S) = 0.14 / (2 x 1 + 32) =0.0041 mol

n(O₂) = m(O₂) / M_W(O₂) = 0.080 / 32 = 0.0025 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 : 1.64 : 1

As we can see the amount ofH₂S is not enough, that’s why H₂S is limiting reagent.

n(Ag₂S) = n(H₂S) = 0.0041mol

The weight of Ag₂S is

m(Ag₂S) = n(Ag₂S) x M_W(Ag₂S) = 0.0041 x (108 x 2 +32) = 1.017 g

Answer: m(Ag₂S) = 1.017 g