The reaction between aluminum and oxygen can be expressed by the following balanced formula:
[tex]4Al+3O_2\rightarrow2Al_2O_3[/tex]We see that 4 moles of aluminum react with 3 moles of oxygen. This means that the O2 to Al ratio is 3/4. Therefore, the moles of O2 needed will be:
[tex]\begin{gathered} molO_2=givenmolAl\times\frac{3molO_2}{4molAl} \\ molO_2=14.8molAl\times\frac{3molO_{2}}{4molAl}=11.1molO_2 \end{gathered}[/tex]Answer: Are required 11.1 moles of O2 to react with 14.8 mol Al.
