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2 November, 08:12

When a mixture of silver metal and sulfur is heated, silversulfide is formed: 16Ag (s) + S8 (s) - -> 8Ag2S (s) a. What mass of Ag2S is produced from a mixture of2.0 g of Ag and 2.0 g of S8? b. What mass of which reactant is left unreacted?

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  1. 2 November, 08:30
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    Mass of Ag₂S which is produced, 2.29 g

    Mass of reactant in excess (S₈) which is left unreacted, 1.70g

    Explanation:

    The balanced reaction is this:

    16 Ag (s) + S₈ (s) → 8Ag₂S (s)

    Molar mass of sulfur: 2g / 256.48 g/m = 0.00779 moles

    Molar mass of silver: 2 g / 107.87 g/m = 0.0185 moles

    For 16 moles of silver, I need 1 mol of S

    For 0.0185 moles of Ag, I will need (0.0185 / 16) = 0.00116 moles

    If I need 0.00116 moles of S, and I have 0.00779 moles it means, that S is my reactant in excess so the limiting reagent is the Ag.

    Let's verify:

    1 mol of S are needed to make react 16 moles of Ag

    0.00779 moles of S, will need (0.00779.16) = 0.124 moles of Ag

    (I only have 0.0185 moles of Ag)

    So the Ag is the limiting reactant, now we can calculate the mass of formed product:

    16 moles of Ag, produce 8 moles of Ag₂S

    0.0185 moles of Ag will produce (0.0185.8) / 16 = 0.00925 moles of Ag₂S

    To find out the mass, let's multiply moles. molar mass

    0.00925 m. 247.8 g/m = 2.29 g

    Mass of the excess, which is left unreacted:

    0.00779 m - 0.00116m = 0.00667 moles

    0.00667m. 256.48 g/m = 1.70 g
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