Answer:
[tex]n_{O_2}^{leftover}=7.7mol[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to set up the corresponding chemical equation:
[tex]4Al+3O_2\rightarrow 2Al_2O_3[/tex]
In such a way, we calculate the moles of aluminum consumed by 13.0 moles of oxygen in the reaction, by applying the 4:3 mole ratio between them:
[tex]n_{Al}=13.0molO_2*\frac{4molAl}{3molO_2} =17.3molAl[/tex]
This means that Al is actually the limiting reactant and oxygen is in excess, for that reason we calculate the moles of oxygen consumed by 7.0 moles of aluminum:
[tex]n_{O_2}=7.0molAl*\frac{3molO_2}{4molAl} =5.3molO_2[/tex]
Thus, the leftover of oxygen is:
[tex]n_{O_2}^{leftover}=13.0mol-5.3mol\\\\n_{O_2}^{leftover}=7.7mol[/tex]
Whereas all the aluminum is assumed to be consumed.
Regards!
