To determine the theoretical yield of [tex]\( \text{Al}_{2}\text{O}_{3} \)[/tex] in this reaction, we need to consider the concept of the limiting reactant. The limiting reactant is the reactant that is completely consumed first in the reaction, limiting the amount of product that can be formed.
Given the balanced equation:
[tex]\[
4 \text{Al} (s) + 3 \text{O}_{2} (g) \rightarrow 2 \text{Al}_{2}\text{O}_{3} (s)
\][/tex]
We are provided that:
- Aluminum (\text{Al}) can produce [tex]\( 4.7 \, \text{g} \, \text{Al}_{2}\text{O}_{3} \)[/tex].
- Oxygen (\text{O}_{2}) can produce [tex]\( 17 \, \text{g} \, \text{Al}_{2}\text{O}_{3} \)[/tex].
To find the theoretical yield, we identify the limiting reactant by comparing the masses of [tex]\( \text{Al}_{2}\text{O}_{3} \)[/tex] that each reactant can produce. The reactant that produces the smaller amount of [tex]\( \text{Al}_{2}\text{O}_{3} \)[/tex] is the limiting reactant.
In this case:
- [tex]\( \text{Al} \)[/tex] can produce [tex]\( 4.7 \, \text{g} \, \text{Al}_{2}\text{O}_{3} \)[/tex].
- [tex]\( \text{O}_{2} \)[/tex] can produce [tex]\( 17 \, \text{g} \, \text{Al}_{2}\text{O}_{3} \)[/tex].
Clearly, [tex]\( 4.7 \, \text{g} \, \text{Al}_{2}\text{O}_{3} \)[/tex] (from \text{Al}) is less than [tex]\( 17 \, \text{g} \, \text{Al}_{2}\text{O}_{3} \)[/tex] (from \text{O}_{2}). Therefore, aluminum (\text{Al}) is the limiting reactant in this reaction.
The theoretical yield of [tex]\( \text{Al}_{2}\text{O}_{3} \)[/tex] is then determined by the amount of product that can be produced by the limiting reactant.
Thus, the theoretical yield of aluminum oxide ([tex]\( \text{Al}_{2}\text{O}_{3} \)[/tex]) is [tex]\( 4.7 \, \text{g} \)[/tex].
Therefore, the theoretical yield is:
[tex]\[ \boxed{4.7 \, \text{g} \, \text{Al}_{2}\text{O}_{3} } \][/tex]
