Silver tarnishes in the presence of hydrogen sulfide (which smells like rotten e

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

How many grams of silver sulfide can be formed from a mixture of 1.1 g Ag, 0.12 g H2S, and 0.093 g O2?

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

g Silver tarnishes in the presence of hydrogen sulfide (which smells like rotten eggs) and oxygen because of the reaction: 4Ag + 2H2S + O2 ? 2Ag2S + 2H2O How many grams of silver sulfide can be formed from a mixture of 1.1 g Ag, 0.12 g H2S, and 0.093 g O2?

Explanation / Answer

molar mass Ag = 107.8682 g/mol

molar mass of H2S = 34.0809 g/mol

molar mass of Ag2S = 247.8 g/mol

molar mass of H2O = 18.01528 g/mol

molar mass of O2 = 32 g/mol

now,

moles = mass/molar mass

so,

moles of Ag = 1.1/107.8682 = 0.0102 moles

moles of H2S = 0.12/34.0809 = 0.003521 moles

moles of O2 = 0.093/32 = 0.00291 moles

we have

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

4 moles of Ag reacts with 2 moles of H2S and 1 mole of O2

so,

H2S is limiting reagent .

as from equation 2 moles of H2S produces 2 moles Ag2S

0.003521 moles of H2S will produce 0.003521 moles of Ag2S

so,

mass of Ag2S produced = moles*molar mass = 0.003521*247.8 = 0.8725 g

2)

from equation ..

4 moles of Ag reacts with 2 moles of H2S ,

so,

0.0102 moles of Ag reacts with 0.0102/2=0.0051 moles of H2S

so,

mass of H2S = 0.0051*34.0809 = 0.1738 g

bt we already have 0.12 g of H2S

so,

additional mass of H2S needed = 0.1738-0.12 = 0.054 g

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