Anonymous asked in Science & MathematicsChemistry · 7 years ago

Consider the unbalanced equation... Ag(s) + H2S(g) + O2(g) Ag2S(s) + H2O(l)?

What is the maximum mass of Ag2S that can be formed when 0.600 g of Ag, 0.190 g H2S and 0.395 g O2 react?

2 Answers

  • 7 years ago
    Favorite Answer

    4 Ag(s) + 2 H2S(g) + O2(g) ==> 2 Ag2S(s) + 2 H2O(l)

    0.600 g Ag / 107.87 g/mole Ag = 0.00556225 moles Ag

    0.190 g H2S / 34.09 g/mole H2S = 0.005573481 moles H2S

    0.395 g O2 / 32.00 g/mole O2 = 0.01234375 moles O2

    Limiting Reagent Time


    0.00556225 moles Ag x 2/4 molar ratio = 0.002781125 moles Ag2S produced

    0.005573481 moles H2S x 2/2 molar ratio = 0.005573481 moles Ag2S produced

    0.01234375 moles O2 x 2/1 molar ratio = 0.0246875 moles Ag2S produced

    Ag is the limiting reagent

    0.00556225 moles Ag x molar ratio of 2 moles Ag2S/4 moles Ag = 0.002781125 moles Ag2S

    0.002781125 moles Ag2S x molar mass of Ag2S (247.81 g/mole) = 0.689190586 g of Ag2S

    0.689 g Ag2S produced rounded to 3 significant figures

  • coxjr
    Lv 4
    3 years ago

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