Question

3. Use the chemical equation to answer the question.
2Ag(s) + H2S(g) → Ag2S(s) + H2(g)
The molar mass of silver (Ag) is 108 g/mol. The molar mass of sulfur (S) is 32 g/mol. The reaction uses 0.04 mol of silver. Which steps show how to determine the mass of silver sulfide (Ag2S) produced in the reaction?

A 2(108 g/mol)+32 g/mol=248 g/mol2(108 g/mol)+32 g/mol=248 g/mol; (248 g/mol)(0.02 mol)=4.96 g(248 g/mol)(0.02 mol)=4.96 g
B 108 g/mol+2(32 g/mol)=172 g/mol108 g/mol+2(32 g/mol)=172 g/mol; (172 g/mol)(0.04 mol)=6.88 g(172 g/mol)(0.04 mol)=6.88 g
C 108 g/mol+2(32 g/mol)=172 g/mol108 g/mol+2(32 g/mol)=172 g/mol; (172 g/mol)(0.02 mol)=3.44 g(172 g/mol)(0.02 mol)=3.44 g
D 2(108 g/mol)+32 g/mol=248 g/mol2(108 g/mol)+32 g/mol=248 g/mol; (248 g/mol)(0.04 mol)=9.92 g

Answer 1

The mass of silver sulfide that would be produced will be 4.96 g and the correct step would be option A.

Stoichiometric equation

First, get the balanced equation of the reaction:

2Ag(s) + H2S(g) → Ag2S(s) + H2(g)

The mole ratio of Ag and Ag2S is 2:1

For 0.04 mole Ag, the equivalent mole of Ag2S would be: 0.04/2 = 0.02 moles.

Mass of 0.02 mole Ag2S = mole x molar mass

Molar mass of Ag2S = 108x2 + 32 = 248

Mass of Ag2S = 0.02 x 248 = 4.96 g

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