To determine the mass of [tex]\( O_2 \)[/tex] needed to form 3.80 moles of [tex]\( Al_2O_3 \)[/tex], we can follow these detailed steps:
1. Understanding the Mole Ratio:
Based on the balanced chemical equation:
[tex]\[
4 \, \text{Al} + 3 \, \text{O}_2 \rightarrow 2 \, \text{Al}_2\text{O}_3
\][/tex]
This equation tells us that 3 moles of [tex]\( O_2 \)[/tex] react to form 2 moles of [tex]\( Al_2O_3 \)[/tex].
2. Calculate the Moles of [tex]\( O_2 \)[/tex] Required:
We need to find out how many moles of [tex]\( O_2 \)[/tex] are required to produce 3.80 moles of [tex]\( Al_2O_3 \)[/tex]:
[tex]\[
\text{To produce } 2 \text{ moles of } Al_2O_3, \text{ we need } 3 \text{ moles of } O_2
\][/tex]
Using the stoichiometric coefficients to find moles of [tex]\( O_2 \)[/tex] needed:
[tex]\[
\text{Moles of } O_2 = \left( \frac{3}{2} \right) \times 3.80 = 5.70 \text{ moles of } O_2
\][/tex]
3. Determine the Mass of [tex]\( O_2 \)[/tex] Needed:
The molar mass of [tex]\( O_2 \)[/tex] is given as [tex]\( 32.0 \, \text{g/mol} \)[/tex]. To find the mass in grams, we multiply the moles of [tex]\( O_2 \)[/tex] needed by the molar mass:
[tex]\[
\text{Mass of } O_2 = 5.70 \, \text{moles} \times 32.0 \, \text{g/mol} = 182.4 \, \text{g}
\][/tex]
Therefore, the mass of [tex]\( O_2 \)[/tex] required to react and form 3.80 moles of [tex]\( Al_2O_3 \)[/tex] is 182 grams.
