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

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Answer:


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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User ThiagoAM
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