# How many chlorine atoms are contained in a 2*mol quantity of FeCl_3?

Aug 18, 2017

$3.6 \cdot {10}^{24}$ Chlorine atoms.

#### Explanation:

Each $F e C {l}_{3}$ has three Chlorines.
Here, we can use a conversion factor.
(3Cl/molecule)*(6.022*10^23 molecules/mol)*(2.0mol)
Molecules and molecules cancel out. Moles and moles cancel out.
We are left with:
$3 C l \cdot 6.022 \cdot {10}^{23} \cdot 2.0$
Simplifying and giving the answer to 2 significant figures, we get:
$3.6 \cdot {10}^{24}$ Chlorine atoms.

Aug 18, 2017

$6 \times {N}_{A}$ chlorine atoms, where ${N}_{A} = 6.022 \times {10}^{23}$

#### Explanation:

If you were asked how many chlorine atoms there were in $\text{1 dozen}$ $F e C {l}_{3}$ formula units, I think you would very quickly answer that there were $\text{3 dozen}$, i.e. $36$ chlorine atoms.

Here you were asked the SAME QUESTION using the mole, ${N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$ as your counting unit.

And thus there are ................

$2 \cdot m o l \times 3 \cdot \text{chlorine atoms/formula unit} \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$

=??*"chlorine atoms".