# If a 25.6*g mass of "octane" is completely combusted, what mass of carbon dioxide will result?

Jun 29, 2017

Approx...$1.8 \cdot m o l \ldots \ldots \ldots .$

#### Explanation:

We need (i) a stoichiometrically balanced equation.....

${C}_{8} {H}_{18} \left(l\right) + \frac{25}{2} {O}_{2} \left(g\right) \rightarrow 8 C {O}_{2} \left(g\right) + 9 {H}_{2} O \left(l\right)$

Is this balanced with respect to mass and charge? It must be if we purport to represent chemical reality.

And (ii) we need equivalent quantities of octane.....

$\text{Moles of octane} = \frac{25.6 \cdot g}{114.23 \cdot g \cdot m o {l}^{-} 1} = 0.224 \cdot m o l$

And given the equation, CLEARLY we KNOW that 8 equiv of carbon dioxide result per equiv of octane.....

And thus a molar quantity of $8 \times 0.224 \cdot m o l$ $C {O}_{2} \left(g\right)$ $=$ ??*mol $=$ ??*g.

Jun 29, 2017

79 grams of carbondioxide

#### Explanation:

${C}_{8} {H}_{18} + 12.5 {O}_{2} \to 8 C {O}_{2} + 9 {H}_{2} O$

This is the chemical reaction.

1 mole of octane mass is 114 grams.

It means if you have 114 grams of octane, you will have 352 grams of $C {O}_{2}$ after the reaction.

However you have only 25.6 grams of octane. You can get how much carbondioxide after the reaction.

$= \frac{25.6 \times 352}{114}$

$= 79$ grams