To determine the mass of silver sulfide (AgS) produced in the reaction using 0.04 mol of silver (Ag), we first need to examine the stoichiometry of the balanced chemical equation.
From the reaction: \[ 2 \text{Ag}(s) + \text{H}_2\text{S}(g) \rightarrow \text{Ag}_2\text{S}(s) + \text{H}_2(g) \]
This indicates that 2 moles of silver produce 1 mole of silver sulfide (Ag₂S). Therefore, if we start with 0.04 mol of silver, we can determine how much silver sulfide will be produced.
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Calculate moles of Ag₂S produced: From the stoichiometry of the reaction: \[ \text{Moles of Ag}_2\text{S} = \frac{0.04 \text{ mol Ag}}{2} = 0.02 \text{ mol Ag}_2\text{S} \]
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Calculate the molar mass of Ag₂S: The molar mass of silver sulfide (Ag₂S) can be calculated as follows: \[ \text{Molar mass of Ag}_2\text{S} = 2 (108 , \text{g/mol}) + 32 , \text{g/mol} = 248 , \text{g/mol} \]
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Calculate the mass of Ag₂S produced: Using the moles of Ag₂S produced and its molar mass: \[ \text{Mass of Ag}_2\text{S} = 0.02 , \text{mol} \times 248 , \text{g/mol} = 4.96 , \text{g} \]
Now, regarding the options provided, the correct steps to determine the mass of silver sulfide produced are:
- Step: 2 (108 g/mol) + 32 g/mol = 248 g/mol; (248 g/mol)(0.02 mol) = 4.96 g
This option correctly calculates the molar mass of silver sulfide and then determines the mass produced from 0.02 moles.
So the correct answer is: 2 (108 g/mol) + 32 g/mol = 248 g/mol; (248 g/mol)(0.02 mol) = 4.96 g.