# How many O_2 molecules are needed to react with 7.87 g of S?

Aug 18, 2016

#### Answer:

Wee assume formation of the $S \left(I V\right)$ oxide, $S {O}_{2} \left(g\right)$

#### Explanation:

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

Moles of $S$ $=$ $\frac{7.87 \cdot g}{32.06 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.245 \cdot m o l$

And thus, by reaction stoichiometry, we need $0.245 \cdot m o l$ dioxygen gas.

So we need, $0.245 \cdot m o l \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1 \text{ oxygen molecules}$.

How many oxygen atoms does this represent?