Answer:
[tex]n_{O_2}=3.84molO_2[/tex]
Explanation:
Hello!
In this case, since the combustion reaction of methanol is:
[tex]CH_3OH+\frac{3}{2} O_2\rightarrow CO_2+2H_2O[/tex]
In such a way, since there is 1:3/2 mole ratio between methanol and oxygen, we can compute the moles of oxygen that are needed to burn 2.56 moles of methanol as shown below:
[tex]n_{O_2}=2.56molCH_3OH*\frac{\frac{3}{2}molO_2}{1molCH_3OH} \\\\n_{O_2}=3.84molO_2[/tex]
Best regards!
