# How many oxygen molecules are required to oxidize a 4.42*g mass of sulfur?

Dec 23, 2017

We address the stoichiometric equation...

$S \left(s\right) + {O}_{2} \left(g\right) \rightarrow S {O}_{2} \left(g\right)$

#### Explanation:

Sulfur COULD be oxidized further, but we will stick with this formula for this calculation.

$\text{Moles of sulfur} = \frac{4.42 \cdot g}{32.06 \cdot g \cdot m o {l}^{-} 1} = 0.138 \cdot m o l$

Given the stoichiometry of the reaction we need $0.138 \cdot m o l$ dioxygen molecules for equivalence, i.e. 0.138*molxx6.022xx10^23*mol^-1=?? ${O}_{2}$ $\text{molecules?}$

What mass of dioxygen does this represent?