Answer:
[tex]\boxed {\boxed {\sf \frac {1}{3} \ mol \ Al_2O_3 \approx 0.34 \ mol \ Al_2O_3}}[/tex]
Explanation:
We will use stoichiometry to solve this problem. The reaction given has a formula of
[tex]4Al+3O_2 \rightarrow 2Al_2O_3[/tex]
The coefficients tell us the number of moles necessary for the reaction.
The reaction requires 3 moles of oxygen to produce 2 moles of aluminum oxide. We can make a ratio.
[tex]\frac {3 \ mol \ O_2}{ 2\ mol \ Al_2O_3}}[/tex]
Since we are doing the reaction with 0.5 moles of oxygen, we multiply the ratio by that number.
[tex]0.5 \ mol \ O_2 *\frac {3 \ mol \ O_2}{ 2 \ mol \ Al_2O_3}}[/tex]
Flip the ratio so the moles of oxygen cancel each other out.
[tex]0.5 \ mol \ O_2 *\frac {2 \ mol \ Al_2O_3}{ 3 \ mol \ O_2}}[/tex]
[tex]0.5 *\frac {2 \ mol \ Al_2O_3}{ 3 }[/tex]
[tex]\frac {1 \ mol \ Al_2O_3}{ 3 }[/tex]
[tex]\frac {1}{3} \ mol \ Al_2O_3 = 0.33333 \ mol \ Al_2O_3 \approx 0.34 \ mol \ Al_2O_3[/tex]
0.5 moles of oxygen produces 1/3 or approximately 0.34 moles of aluminum oxide.
