# A 15*g mass of methane is combusted with a 50*g mass of dioxygen gas. Will combustion be complete?

Nov 27, 2016

$C {H}_{4} + 2 {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(g\right)$

#### Explanation:

You have the balanced chemical equation. Not only does this tell you the mass equivalence, i.e. how many grams of methane react with how many grams of dioxygen, you can include parameters that tell you how energy is transferred in the the reaction.

Here, there are $\frac{15.0 \cdot g}{16.04 \cdot g \cdot m o {l}^{-} 1} \cong 1 \cdot m o l$ $\text{methane}$.

And for complete combustion, this requires $2 \times \frac{15.0 \cdot g}{16.04 \cdot g \cdot m o {l}^{-} 1} \times 32.00 \cdot g \cdot m o {l}^{-} 1 \cong 68 \cdot g$ dioxygen gas.

You thus have an insufficient mass of dioxygen for complete combustion. Complete combustion requires the given mass.

And so we don't know how much carbon dioxide will result; certainly it will be much less than the stoichiometric amount.