Question #25715
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?
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?
Expert's answer
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) = 0.14 / (2 x 1 + 32) =0.0041 mol
n(O2) = m(O2) / MW(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 MW(Ag2S) = 0.0041 x (108 x 2 +32) = 1.017 g
Answer: m(Ag2S) = 1.017 g
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) = 0.14 / (2 x 1 + 32) =0.0041 mol
n(O2) = m(O2) / MW(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 MW(Ag2S) = 0.0041 x (108 x 2 +32) = 1.017 g
Answer: m(Ag2S) = 1.017 g
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#339914 on Dec 2023
#339914 on Dec 2023