Question

How many miles of O2 are needed to burn 1.45 mil of C8H18

Answer 1

`n_(O_2)=18.1molO_2`

Step-by-step explanation:

Hello there!

In this case, according to the combustion of octane:

`C_8H_1_8+25/2O_2\rightarrow 8CO_2+9H_2O`

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:

`n_(O_2)=1.45molC_8H_1_8*(25/2molO_2)/(1molC_8H_1_8) \\\\n_(O_2)=18.1molO_2`

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