This is a stoichiometry problem. Follow this four-step procedure to work all stoichiometry problems. If it is a limiting reagent problem it will work those, too, AFTER the limiting reagent has been determined. First off this is not a very practical problem; however, let's let that go and just solve the problem as if it can be easily done.
Step 1. Write and balance the equation.
2Ag(s) + H2S(g) ==> Ag2S + H2
Step 2. Determine moles of what you have. That usually is moles = grams/molar mass; however, that is already done and the problem tells you mols Ag = 0.04 moles.
Step 3. Using the coefficients in the balanced equation, convert mols of what you have (in this case it is moles Ag) into moles of what you want (in this case you want Ag2S).
0.04 mols Ag x (1 mol Ag2S/2 mols Ag) = 0.04 x 1/2 = 0.02 mols Ag2S
Step 4. Convert moles of what you want to grams (usually) or whatever.
grams Ag2S = moles Ag2S x molar mass Ag2S. The molar mass Ag2S = 2*108 + 32 = 248. Can you take this and fit it into one of the answers. My first suggestion is to work the problem using my steps and use that answer to eliminate wrong answers from the choices. I'll check back Stay in tough if you still have questions.
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?
108 g/mol+2(32 g/mol)=172 g/mol; (172 g/mol)(0.04 mol)=6.88 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)(1 mol Ag2S / 2mol Ag)=4.96 g
108 g/mol+2(32 g/mol)=172 g/mol; (172 g/mol)(0.04 mol)(1 mol Ag2S/
2molAg)=3.44 g
I honestly don’t even know where to start.
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