To determine how many moles of oxygen gas [tex]\( O_2 \)[/tex] are needed to react completely with 54.0 g of aluminum, we need to follow a step-by-step conversion and use stoichiometry based on the chemical equation:
[tex]\[
4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3
\][/tex]
### Step-by-Step Solution:
1. Convert the mass of aluminum to moles:
We start with 54.0 grams of aluminum (Al).
The molar mass of aluminum (Al) is [tex]\[ 26.98 \text{ grams per mole (g/mol)}. \][/tex]
To convert grams to moles, we use the formula:
[tex]\[
\text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}}
\][/tex]
Substituting in the given values:
[tex]\[
\text{moles of Al} = \frac{54.0 \text{ g}}{26.98 \text{ g/mol}} = 2.001482579688658 \text{ mol}
\][/tex]
2. Use the stoichiometric ratio to find moles of [tex]\( O_2 \)[/tex]:
The balanced chemical equation shows that 4 moles of Al react with 3 moles of [tex]\( O_2 \)[/tex]:
[tex]\[
4 \text{ Al} + 3 \text{ O}_2 \rightarrow 2 \text{ Al}_2\text{O}_3
\][/tex]
Therefore, the molar ratio of [tex]\( O_2 \)[/tex] to Al is:
[tex]\[
\frac{3 \text{ mol } O_2}{4 \text{ mol } Al}
\][/tex]
To find the moles of [tex]\( O_2 \)[/tex] needed, we multiply the moles of Al by the stoichiometric ratio:
[tex]\[
\text{moles of } O_2 = \left( 2.001482579688658 \text{ mol Al} \right) \times \left( \frac{3 \text{ mol } O_2}{4 \text{ mol } Al} \right) = 1.5011119347664936 \text{ mol } O_2
\][/tex]
### Green Box Value:
The value in the green box for the conversion of 54.0 g Al is:
[tex]\[
\boxed{1 \text{ mol Al}}
\][/tex]
This value represents the conversion factor in the stoichiometric calculation. Thus, 54.0 g of aluminum is equivalent to 2.001482579688658 moles of aluminum.
### Final Answer:
The moles of oxygen gas [tex]\( \text{O}_2 \)[/tex] needed to react with 54.0 g of aluminum is:
[tex]\[
1.5011119347664936 \text{ mol } O_2
\][/tex]
