Use the chemical equation to answer the question.

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?

(1 point)
Responses

2(108 g/mol)+32 g/mol=248 g/mol
; (248 g/mol)(0.02 mol)=4.96 g
2(108 g/mol)+32 g/mol=248 g/mol
; (248 g/mol)(0.02 mol)=4.96 g

2(108 g/mol)+32 g/mol=248 g/mol
; (248 g/mol)(0.04 mol)=9.92 g
2(108 g/mol)+32 g/mol=248 g/mol
; (248 g/mol)(0.04 mol)=9.92 g

108 g/mol+2(32 g/mol)=172 g/mol
; (172 g/mol)(0.02 mol)=3.44 g
108 g/mol+2(32 g/mol)=172 g/mol
; (172 g/mol)(0.02 mol)=3.44 g

108 g/mol+2(32 g/mol)=172 g/mol
; (172 g/mol)(0.04 mol)=6.88 g

Answer 1

To determine the mass of silver sulfide (Ag2S) produced from the reaction given that 0.04 mol of silver (Ag) is used, you need to follow these steps:

1. Calculate the molar mass of silver sulfide (Ag2S): \[ \text{Molar mass of Ag2S} = 2 \times \text{molar mass of Ag} + \text{molar mass of S} = 2(108 , \text{g/mol}) + 32 , \text{g/mol} = 248 , \text{g/mol} \]

2. Determine the amount of silver sulfide produced:

   • From the balanced equation, 2 moles of Ag produce 1 mole of Ag2S.
   • Therefore, if 0.04 mol of Ag is used, the amount of Ag2S produced will be: \[ \text{Moles of Ag2S} = \frac{0.04 , \text{mol Ag}}{2} = 0.02 , \text{mol Ag2S} \]

3. Calculate the mass of silver sulfide produced: \[ \text{Mass of Ag2S} = \text{molar mass of Ag2S} \times \text{moles of Ag2S} = 248 , \text{g/mol} \times 0.02 , \text{mol} = 4.96 , \text{g} \]

Thus, the correct response that shows these steps is:

\[ 2(108 , \text{g/mol}) + 32 , \text{g/mol} = 248 , \text{g/mol}; \quad (248 , \text{g/mol})(0.02 , \text{mol}) = 4.96 , \text{g} \]

This response involves calculating the molar mass of Ag2S and then using it to find the mass based on the moles of Ag2S produced.