Use the chemical equation to answer the question.

Question

2Ag(s) + H2S(g) → Ag 2S(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
(Ag S) produced in the reaction?
(1 point)
2 2 2
2
108 g/mol + 2 (32 g/mol) = 172 g/mol; (172 g/mol)(0.02 mol) = 3.44 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)(0.02 mol) = 4.96 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 in the reaction, we need to first find out how many moles of Ag2S are produced based on the moles of silver (Ag) used in the reaction.

From the balanced chemical equation:

$$2Ag(s)+H2S(g)\to Ag2S(s)+H2(g)$$

We see that 2 moles of silver yield 1 mole of silver sulfide. If 0.04 moles of silver are used, we can determine the moles of silver sulfide produced:

1. Calculate the moles of Ag2S produced: $$\text{moles of Ag2S}=\frac{0.04,\text{mol Ag}}{2}=0.02,\text{mol Ag2S}$$

2. Now, we need to calculate the molar mass of Ag2S: $$\text{molar mass of Ag2S}=2\times (108,\text{g/mol})+(32,\text{g/mol})=248,\text{g/mol}$$

3. Finally, we calculate the mass of Ag2S produced using the moles of Ag2S and its molar mass: $$\text{mass of Ag2S}=(248,\text{g/mol})(0.02,\text{mol})=4.96,\text{g}$$

Thus, the correct steps that show how to determine the mass of silver sulfide produced are:

Step:

• $2(108,\text{g/mol})+32,\text{g/mol}=248,\text{g/mol};,(248,\text{g/mol})(0.02,\text{mol})=4.96,\text{g}$

This matches with the third option listed:

3. $2(108,\text{g/mol})+32,\text{g/mol}=248,\text{g/mol};(248,\text{g/mol})(0.02,\text{mol})=4.96,\text{g}$