Answer :
The percent yield for the reaction is approximately 31.7%, indicating that only about one-third of the expected amount of ZnO was obtained in the experiment. This lower yield may be due to factors like incomplete reactions or experimental errors.
Step 1: Calculate the theoretical yield of ZnO:
1. Find the moles of ZnS:
[tex]\[ \text{moles of ZnS} = \frac{46.5 \, \text{g}}{molar \, mass \, of \, ZnS} = \frac{46.5 \, \text{g}}{65.38 \, \text{g/mol}} \approx 0.711 \, \text{mol} \][/tex]
2. Use the mole ratio to find moles of ZnO:
[tex]\[ \text{moles of ZnO} = \text{moles of ZnS} = 0.711 \, \text{mol} \][/tex]
3. Convert moles to grams:
[tex]\[ \text{mass of ZnO} = \text{moles of ZnO} \times molar \, mass \, of \, ZnO = 0.711 \, \text{mol} \times 81.38 \, \text{g/mol} \approx 58.0 \, \text{g} \][/tex]
Step 2: Calculate the percent yield:
[tex]\[ \text{Percent Yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100 \][/tex]
[tex]\[ \text{Percent Yield} = \left( \frac{18.4 \, \text{g}}{58.0 \, \text{g}} \right) \times 100 \approx 31.7\% \][/tex]
The percent yield for the reaction is approximately 31.7%. This means that 31.7% of the expected amount of ZnO was obtained in the actual experiment. The lower-than-expected yield could be attributed to various factors, such as incomplete reactions, side reactions, or experimental errors.
The question probable maybe:
Determine the percent yield for the reaction between 46.5 grams of ZnS and 20.42 grams of O₂ if 18.4 grams of ZnO is recovered along with an unknown quantity of sulfur dioxide.
ZnS + O₂ -> ZnS + SO₂