# Elemental sulfur occurs as octatomic molecules, S_8. What mass (g) of fluorine gas is needed to react completely with 17.8 grams of sulfur to form sulfur hexafluoride?

Feb 21, 2016

$\text{63.2 g F"_2}$ are needed to react completely with $\text{17.8 g S"_8}$ to produce $\text{SF"_6}$.

#### Explanation:

${\text{S"_8+"24F}}_{2}$$\rightarrow$$\text{8SF"_6}$

Then determine the molar masses of $\text{S"_8}$ and $\text{F"_2}$ by multiplying the subscript of each element by its atomic mass from the periodic table in g/mol.

$\text{S"_8} :$ (8xx32.06 "g/mol")="256.48 g/mol"
$\text{F"_2} :$ (2xx18.998 "g/mol")="37.996 g/mol"

Determine the mass of $\text{F"_2}$ needed to react completely with $\text{S"_8}$ to form $\text{SF"_6}$.

1. Divide the given mass $\text{S"_8}$ by its molar mass.
2. Multiply the mole ratio between $\text{F"_2}$ and $\text{S"_8}$ from the balanced equation with the moles $\text{F"_2}$ in the numerator: $\left(24 {\text{mol F"_2)/(1"mol S}}_{8}\right)$
3. Multiply moles $\text{F"_2}$ times its molar mass to get grams $\text{F"_2}$.

$17.8 {\cancel{\text{g S"_8xx(1cancel"mol S"_8)/(256.48cancel"g S"_8)xx(24cancel"mol F"_2)/(1cancel"mol S"_8)xx(37.996"g F"_2)/(1cancel"mol F"_2)="63.2 g F}}}_{2}$ rounded to three significant figures