# What mass of benzene is required for a combustion reaction that yields a mass of 42*g with respect to carbon dioxide?

Sep 25, 2016

Approx. $13 \cdot g$ benzene were required.

#### Explanation:

Given the stoichiometric rxn:

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

As a product we have $\frac{42.0 \cdot g}{44.01 \cdot g \cdot m o {l}^{-} 1}$ $\cong$ $1 \cdot m o l \text{ carbon dioxide}$ gas.

And thus approx. $\frac{1}{6} \cdot m o l$ benzene were required for the combustion, $=$ $\frac{1}{6} m o l \times 78.0 \cdot g \cdot m o {l}^{-} 1$ $=$ ??*g.

You can of course make a better estimate than I did.