# What mass of oxygen is needed for the complete combustion of 8.90 * 10^-3 g of methane?

Feb 28, 2016

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

#### Explanation:

$\text{Moles of methane}$ $=$ $\frac{8.90 \times {10}^{-} 3 \cdot g}{16.04 \cdot g \cdot m o {l}^{-} 1}$ $=$ ?? $\text{moles}$

Given the stoichiometric equation, quite clearly, we need twice this molar quantity of dioxygen gas for complete combustion:

$2 \times \frac{8.90 \times {10}^{-} 3 \cdot g}{16.04 \cdot g \cdot m o {l}^{-} 1} \times 32.00 \cdot g \cdot m o {l}^{-} 1 \text{ dioxygen gas}$

$=$ ??g $\text{dioxygen gas}$.