To determine how many moles of oxygen gas [tex]\(\left( O_2 \right)\)[/tex] are required to react with [tex]\(54.0 \, \text{g}\)[/tex] of aluminum using the balanced equation:
[tex]\[4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2 \text{O}_3,\][/tex]
we need to follow these steps:
1. Convert grams of Aluminum to moles of Aluminum:
The molar mass of aluminum (Al) is needed for this conversion.
Given that [tex]\(1\)[/tex] mole of Aluminum has a mass of [tex]\(26.98 \, \text{g}\)[/tex], the conversion factor related to the mass of Aluminum should be:
[tex]\[
\frac{1 \, \text{mol Al}}{26.98 \, \text{g Al}}
\][/tex]
2. Identify the correct conversion factor:
Based on the choices:
- Choice 1: [tex]\(4 \, \text{mol Al}\)[/tex]
- Choice 2: [tex]\(3 \, \text{mol O}_2\)[/tex]
- Choice 3: [tex]\(26.98 \, \text{g Al}\)[/tex]
The correct molar mass of Aluminum is [tex]\(26.98 \, \text{g Al}\)[/tex], so the answer choice number for this conversion factor should be:
__Choice 3: \(26.98 \, \text{g Al}__
