Question 09624

Sep 5, 2016

${\text{7.95 g O}}_{2}$

Explanation:

Start by writing the balanced chemical equation that describes this combustion reaction

$\textcolor{b l u e}{2} {\text{C"_ 8"H"_ (18(l)) + color(red)(25)"O"_ (2(g)) -> 16"CO"_ (2(g)) + 18"H"_ 2"O}}_{\left(g\right)}$

Notice that the reaction consumes $\textcolor{red}{25}$ moles of oxygen gas for every $\textcolor{b l u e}{2}$ moles of octane that take part in the reaction.

In other words, regardless of how many moles of octane take part in the reaction, you will always have

$\text{moles of octane"/"moles of oxygen gas} = \frac{\textcolor{b l u e}{2}}{\textcolor{red}{25}}$

To find the mass of oxygen gas needed for the reaction of $\text{2.27 g}$ of octane, convert the mass of octane to moles by using the compound's molar mass

2.27 color(red)(cancel(color(black)("g"))) * ("1 mole C"_8"H"_18)/(114.23 color(red)(cancel(color(black)("g")))) = "0.01987 moles C"_8"H"_18

Now use the aforementioned mole ratio to find the number of moles of oxygen gas needed for the reaction

0.01987 color(red)(cancel(color(black)("moles C"_8"H"_18))) * (color(red)(25)color(white)(a)"moles O"_2)/(color(blue)(2)color(red)(cancel(color(black)("moles C"_8"H"_18)))) = "0.2484 moles O"_2

To convert the number of moles to grams, use the molar mass of oxygen gas

0.2484color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(green)(bar(ul(|color(white)(a/a)color(black)("7.95 g")color(white)(a/a)|)))#

The answer is rounded to three sig figs.