Answer:
[tex]n_{O_2}=18.1molO_2[/tex]
Explanation:
Hello there!
In this case, according to the combustion of octane:
[tex]C_8H_1_8+25/2O_2\rightarrow 8CO_2+9H_2O[/tex]
We can see there is a 1:25/2 mole ratio of octane to oxygen; therefore, we can calculate the moles of oxygen via the following stoichiometric factor:
[tex]n_{O_2}=1.45molC_8H_1_8*\frac{25/2molO_2}{1molC_8H_1_8} \\\\n_{O_2}=18.1molO_2[/tex]
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