High School

8. Zinc sulfide and oxygen gas react to form zinc oxide and sulfur dioxide. Determine the mass of ZnO that should be produced in a reaction between 46.5 g of ZnS and excess oxygen.

Answer :

To solve this problem, we need to understand the chemical reaction taking place and apply stoichiometry to find the mass of zinc oxide (ZnO) produced from a given amount of zinc sulfide (ZnS).

Chemical Reaction:

The reaction between zinc sulfide and oxygen can be represented by the balanced chemical equation:

[tex]2\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2[/tex]

Step-by-Step Calculation:

  1. Determine the Molar Masses:

    • Molar mass of ZnS (Zn: 65.38 g/mol, S: 32.07 g/mol):
      [tex]\text{Molar mass of ZnS} = 65.38 + 32.07 = 97.45 \text{ g/mol}[/tex]
    • Molar mass of ZnO (Zn: 65.38 g/mol, O: 16.00 g/mol):
      [tex]\text{Molar mass of ZnO} = 65.38 + 16.00 = 81.38 \text{ g/mol}[/tex]

  2. Convert Mass of ZnS to Moles:

    • Given mass of ZnS: 46.5 g
    • Moles of ZnS:
      [tex]\text{Moles of ZnS} = \frac{46.5 \text{ g}}{97.45 \text{ g/mol}} \approx 0.477 \text{ moles}[/tex]

  3. Use the Stoichiometry of the Reaction:

    • According to the balanced equation, 2 moles of ZnS produce 2 moles of ZnO.
    • Therefore, moles of ZnO produced = moles of ZnS = 0.477 moles

  4. Convert Moles of ZnO to Mass:

    • Mass of ZnO produced:
      [tex]\text{Mass of ZnO} = 0.477 \text{ moles} \times 81.38 \text{ g/mol} \approx 38.83 \text{ g}[/tex]

Therefore, approximately 38.83 grams of ZnO should be produced in the reaction when 46.5 grams of ZnS react with excess oxygen.

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