To find the number of moles of oxygen gas ([tex]\(O_2\)[/tex]) needed to react with 54.0 grams of aluminum ([tex]\(Al\)[/tex]), we can follow these steps:
1. Calculate the number of moles of aluminum:
First, we'll convert the mass of aluminum to moles using the molar mass of aluminum. The molar mass of [tex]\(\text{Al}\)[/tex] is 26.98 g/mol.
[tex]\[
\text{Moles of Aluminum} = \frac{\text{Mass of Aluminum}}{\text{Molar Mass of Aluminum}} = \frac{54.0 \text{ g}}{26.98 \text{ g/mol}}
\][/tex]
It comes out to approximately 2.0015 moles of aluminum.
2. Determine the mole ratio from the balanced chemical equation:
The balanced chemical equation is:
[tex]\[
4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3
\][/tex]
From this equation, we see that 4 moles of [tex]\(\text{Al}\)[/tex] react with 3 moles of [tex]\(\text{O}_2\)[/tex].
Therefore, the mole ratio of [tex]\(\text{Al}\)[/tex] to [tex]\(\text{O}_2\)[/tex] is:
[tex]\[
\frac{3 \text{ moles} \, \text{O}_2}{4 \text{ moles} \, \text{Al}} = 0.75
\][/tex]
3. Calculate the moles of oxygen gas needed:
Using the number of moles of aluminum and the mole ratio, we can find the moles of [tex]\(\text{O}_2\)[/tex] needed.
[tex]\[
\text{Moles of Oxygen Gas} = \text{Moles of Aluminum} \times \text{Mole Ratio} = 2.0015 \text{ moles Al} \times 0.75
\][/tex]
This calculation gives us approximately 1.5011 moles of oxygen gas.
Therefore, 1.5011 moles of oxygen gas ([tex]\(O_2\)[/tex]) are needed to completely react with 54.0 grams of aluminum ([tex]\(\text{Al}\)[/tex]).
To fill in the conversion factors in the given table:
[tex]\[
\begin{tabular}{l|c|c}
54.0 g Al & \boxed{\text{1 mole Al}} & \boxed{\text{26.98 g Al}} \\
\hline & \boxed{\text{3 moles O}_2} & 4 moles Al \\
\end{tabular}
\][/tex]
Here, the green box should contain "26.98 g Al", as it is the molar mass of aluminum.
