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How many moles of oxygen gas [tex]$\left( O_2\right)$[/tex] are needed to completely react with [tex]$54.0 \text{ g}$[/tex] of aluminum?

Given the reaction:
[tex]$
4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3
$[/tex]

Let's set up the conversions first. What goes in the green box?

\begin{tabular}{l|l|l}
[tex]$54.0 \text{ g Al}$[/tex] & {[]} & {[]} \\
\hline & {[tex]$[?]$[/tex]} & {[]}
\end{tabular}

1. [tex]$4 \text{ mol Al}$[/tex]
2. [tex]$3 \text{ mol O}_2$[/tex]
3. [tex]$26.98 \text{ g Al}$[/tex]

Enter the answer choice number.



Answer :

GinnyAnswer
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}__

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