# Question 58872

Mar 29, 2016

${\text{122,000 L O}}_{2}$

#### Explanation:

The idea here is that you need to use the density and volume of octane to determine how many moles you have in that sample.

The balanced chemical equation will then help you find the number of moles of oxygen needed to ensure that all the moles of octane take part in the reaction.

The balanced chemical equation for the combustion of octane, ${\text{C"_8"H}}_{18}$, looks like this

$\textcolor{red}{2} {\text{C"_8"H"_text(18(l]) + color(blue)(25)"O"_text(2(g]) -> 16"CO"_text(2(g]) + 18"H"_2"O}}_{\textrm{\left(l\right]}}$

The $\textcolor{red}{2} : \textcolor{b l u e}{25}$ mole ratio that exists between the two reactants tells you that the reaction will consume $25$ moles of oxygen gas for every $2$ moles of octane that take part in the reaction.

So, use the density and volume of octane to find the mass of the sample

18.5color(red)(cancel(color(black)("gal"))) * (3.79color(blue)(cancel(color(black)("L"))))/(1color(red)(cancel(color(black)("gal")))) * (1000color(green)(cancel(color(black)("mL"))))/(1color(blue)(cancel(color(black)("L")))) * "0.703 g"/(1color(green)(cancel(color(black)("mL")))) = 4.93 * 10^4"g"

Use octane's molar mass to convert this mass to moles

4.93 * 10^4color(red)(cancel(color(black)("g"))) * "1 mole octane"/(114.23color(red)(cancel(color(black)("g")))) = "431.6 moles octane"

According to the aforementioned mole ratio, this many moles of octane would require

431.6color(red)(cancel(color(black)("moles octane"))) * (color(blue)(25)color(white)(a)"moles O"_2)/(color(red)(2)color(red)(cancel(color(black)("moles octane")))) = "5395 moles O"_2

Now, you're working under STP conditions, which imply a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$.

Under these specific conditions for pressure and temperature, one mole of any ideal gas occupies $\text{22.7 L } \to$ this is known as the molar volume of a gas at STP.

In your case, $0.5395$ moles of oxygen gas would occupy

5395color(red)(cancel(color(black)("moles O"_2))) * overbrace("22.7 L"/(1color(red)(cancel(color(black)("mole O"_2)))))^(color(purple)("molar volume of a gas at STP")) = "122,470 L"#

Rounded to three sig figs, the answer will be

${V}_{{O}_{2}} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{122,000 L} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.

SIDE NOTE More often than not, the molar volume of a gas at STP is said to be equal to $\text{22.4 L}$.

This value corresponds to the old definition of STP conditions, which implied a pressure of $\text{1 atm}$ and a temperature of ${0}^{\circ} \text{C}$.

If this is the value given to you, simply redo the last calculation using $\text{22.4 L}$ instead of $\text{22.7 L}$.