Question

if air is 20.9% oxygen by volume, how many liters of air are needed to complete the combustion of 25.0L of octane vapor at STP?

Answer 1

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Answer 2

Answer: 1,495.11 L volume of air will be required for the complete combustion of octane vapors of 25 L.

Explanation:

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

At STP, 1 mole of gas occupies 22.4 L of volume

Then 25.0 L of volume will be occupied by the :$\frac{1}{22.4}\times 25 mol=1.1160 mol$

According to reaction, 2 mole of $2C_8H_{18}$ reacts with 25 moles od $O_2$ , then 1.1160 moles will react with:

$\frac{25}{2}\times 1.1160 mol$ of $O_2$ gas that is 13.95 moles.

At STP, 1 mole of gas occupies 22.4 L of volume

So, 13.95 moles of $O_2$ will occupy the volume:

$=22.4 L\mol\times 13.95 mol=312.48 L$

If air is 20.9% oxygen by volume,then the volume of air required for the complete combustion of octane vapors is:

$20.9\%=\frac{312.48 L}{\text{Volume of air}}\times 100$

Volume of the air = 1,495.11 L

1,495.11 L volume of air will be required for the complete combustion of octane vapors of 25 L.