Calculate the volume of chlorine that is required to convert 1 kg of iron in ferric 3 chloride at standard conditions?

Apr 28, 2016

Approx. $600 \cdot L$ at $\text{STP}$.

Explanation:

$F e \left(s\right) + \frac{3}{2} C {l}_{2} \left(g\right) \rightarrow F e C {l}_{3} \left(s\right)$

Moles of iron $=$ $\frac{1000 \cdot \cancel{g}}{55.85 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ $17.91 \cdot m o l$.

Given the rxn stoichiometry, we need at a bare minimum, $\frac{3}{2}$ this molar quantity of chlorine gas, i.e. $26.9$ $m o l$ $C {l}_{2}$ gas.

Now I know that under standard condtions, $1$ $m o l$ of ideal gas has a volume of $22.4 \cdot L$. (This would be given as supplementary material in any examination!).

Thus $26.9 \cdot \cancel{m o l} \times 22.4 \cdot L \cdot \cancel{m o {l}^{-} 1}$ $=$ ??*L