# What mass of carbon dioxide results from complete combustion of a 9.75*L volume of benzene?

Apr 20, 2017

We need (i) a stoichiometric equation...........

........and we get approx. $29 \cdot k g$ of carbon dioxide.

#### Explanation:

We will be requiring $\left(i\right)$ ${C}_{6} {H}_{6} \left(l\right) + \frac{15}{2} {O}_{2} \left(g\right) \rightarrow 6 C {O}_{2} \left(g\right) + 3 {H}_{2} O \left(l\right)$

$\left(i i\right)$ the density of benzene, ${\rho}_{\text{benzene}} = 0.88 \cdot g \cdot m {L}^{-} 1$,

And $\left(i i i\right)$, equivalent quantities of benzene........

$\text{Moles of benzene} = \frac{9.75 \cdot L \times 0.88 \cdot g \cdot m {L}^{-} 1 \cdot {10}^{3} \cdot m L \cdot {L}^{-} 1}{78.11 \cdot g \cdot m o {l}^{-} 1}$

$= 109.9 \cdot m o l$.

Given the stoichiometry, the resultant mass of carbon dioxide is......$6 \times 109.9 \cdot m o l \times 44.01 \cdot g \cdot m o {l}^{-} 1 \cong 29 \cdot k g$