# If 10.0 g of aluminum sulfide are produced by the reaction of aluminum and sulfur, how many grams or sulfur were needed?

Jun 9, 2018

Approx. $6.5 \cdot g$ of sulfur….

#### Explanation:

We interrogate the stoichiometric reaction....

$2 A l \left(s\right) + 3 S \left(s\right) \stackrel{\Delta}{\rightarrow} A {l}_{2} {S}_{3} \left(s\right)$

Aluminum is oxidized, and sulfur is reduced (note that sometimes, we represent elemental sulfur, i.e. $\text{flowers of sulfur}$, as ${S}_{8}$..)

$\text{Moles of aluminum sulfide} = \frac{10.0 \cdot g}{150.16 \cdot g \cdot m o {l}^{-} 1} = 0.0666 \cdot m o l$...

And thus we need $\text{3 equivs}$ of sulfur oxidant...

-=3xx0.0666*molxx32.06*g*mol^-1=??*g