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:
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]
- Molar mass of ZnS (Zn: 65.38 g/mol, S: 32.07 g/mol):
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]
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
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]
- Mass of ZnO produced:
Therefore, approximately 38.83 grams of ZnO should be produced in the reaction when 46.5 grams of ZnS react with excess oxygen.