Absolutely! Let's solve this step-by-step.
1. Determine the molar masses:
- The molar mass of [tex]\( \text{ZnS} \)[/tex] is 97.46 g/mol.
- The molar mass of [tex]\( \text{O}_2 \)[/tex] is 32.00 g/mol.
2. Convert the mass of [tex]\( \text{ZnS} \)[/tex] to moles:
The given mass of [tex]\( \text{ZnS} \)[/tex] is 0.04 g.
To find the moles of [tex]\( \text{ZnS} \)[/tex]:
[tex]\[
\text{moles of ZnS} = \frac{\text{mass of ZnS}}{\text{molar mass of ZnS}} = \frac{0.04 \, \text{g}}{97.46 \, \text{g/mol}} \approx 0.0004104 \, \text{moles}
\][/tex]
3. Use the stoichiometry of the reaction:
According to the balanced equation:
[tex]\[
2 \, \text{ZnS} + 3 \, \text{O}_2 \rightarrow 2 \, \text{ZnO} + 2 \, \text{SO}_2
\][/tex]
From the coefficients, we see that 2 moles of [tex]\( \text{ZnS} \)[/tex] react with 3 moles of [tex]\( \text{O}_2 \)[/tex].
Therefore, for every 2 moles of [tex]\( \text{ZnS} \)[/tex], 3/2 or 1.5 moles of [tex]\( \text{O}_2 \)[/tex] are required.
So, the moles of [tex]\( \text{O}_2 \)[/tex] needed for [tex]\( 0.0004104 \)[/tex] moles of [tex]\( \text{ZnS} \)[/tex] is:
[tex]\[
\text{moles of } \text{O}_2 = \left( \frac{3}{2} \right) \times \text{moles of ZnS} = 1.5 \times 0.0004104 \approx 0.0006156 \, \text{moles}
\][/tex]
4. Convert moles of [tex]\( \text{O}_2 \)[/tex] to grams:
To find the mass of [tex]\( \text{O}_2 \)[/tex], multiply the moles of [tex]\( \text{O}_2 \)[/tex] by the molar mass of [tex]\( \text{O}_2 \)[/tex]:
[tex]\[
\text{mass of } \text{O}_2 = \text{moles of } \text{O}_2 \times \text{molar mass of } \text{O}_2 = 0.0006156 \, \text{moles} \times 32.00 \, \text{g/mol} \approx 0.0197 \, \text{g}
\][/tex]
Therefore, to consume [tex]\( 0.04 \)[/tex] grams of [tex]\( \text{ZnS} \)[/tex], you need approximately [tex]\( 0.0197 \)[/tex] grams of [tex]\( \text{O}_2 \)[/tex].
