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25 October, 07:30

Zinc (II) sulfide reacts with oxygen according to the reaction: A reaction mixture initially contains 3.0 mol ZnS and 2.0 mol O2. Once the reaction has occurred as completely as possible, what is excess reactant and what amount (in moles) of the excess reactant is left?

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  1. 25 October, 08:34
    0
    Answer: Zinc (II) sulfide (ZnS) is in excess. there will remain 1.67 moles

    Explanation:

    Step 1: Data given

    Zinc (II) sulfide = ZnS

    oxygen = O2

    Number of moles ZnS = 3.0 moles

    Number of moles O2 = 2.0 moles

    Step 2: The balanced equation

    2ZnS + 3O2 → 2ZnO + 2SO2

    Step 3: Calculate the limiting reactant

    For 2 mles zinc (II) sulfide we need 3 moles oxygen to produce 2 moles zinc oxide and 2 moles sulfur dioxide

    O2 is the limiting reactant. There willreact 2.0 moles.

    ZnS is in excess. There will react 2/3*2.0 = 1.33 moles

    There will remain 3.0 - 1.33 = 1.67 moles ZnS

    Step 4: Calculate products

    For 2 mles zinc (II) sulfide we need 3 moles oxygen to produce 2 moles zinc oxide and 2 moles sulfur dioxide

    For 2.0 moles O2 we'll have 1.33 moles ZnO and 1.33 moles SO2

    Answer: Zinc (II) sulfide (ZnS) is in excess. there will remain 1.67 moles
  2. 25 October, 09:47
    0
    Excess reagent is the ZnS. After the reaction is complete, 1.67 moles of sulfide remain.

    Explanation:

    This is the reaction to work with:

    2ZnS + 3O₂ → 2ZnO + 2SO₂

    Limiting reagent is the oxygen. We confirm it by a rule of three

    2 moles of sulfide can react to 3 moles of O₂

    Therefore 3 moles of ZnS will react to (3. 3) / 2 = 4.5 moles (we need 4.5 moles of O₂ and we only have 2 moles, that's why the O₂ is the limiting)

    Excess reagent is the zinc (II) sulfide

    3 moles of oxygen react to 2 moles of ZnS

    2 moles of O₂ will react to (2. 2) / 3 = 1.33 moles of ZnS (it is ok, be cause we have 3 moles, and we need only 1.33)

    After the reaction is complete (3 - 1.33) = 1.67 moles of sulfide remain.
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