# How do we work out the limiting reagent in combustion reactions?

Jul 6, 2017

You will have to supply a few details..........

#### Explanation:

Let's take the combustion of methane........which reaction cooked my dinner tonight.

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

And suppose I used a mass of $10 \cdot g$ of methane...........Now clearly methane is the LIMITING REAGENT, and the dioxygen was in excess. Do you agree?

$\text{Moles of methane} = \frac{10 \cdot g}{16.01 \cdot g \cdot m o {l}^{-} 1} = 0.625 \cdot m o l$

And we require $0.625 \cdot m o l \times 2 = 1.250 \cdot m o l$ $\text{dioxygen gas}$, which represents a mass of how much?