# What mass of chlorine gas is required to react with a 0.567*g mass of phosphorus to make PCl_5?

Sep 26, 2016

A bit over $3 \cdot g$ of $C {l}_{2}$ gas.

#### Explanation:

We need (i) a chemical reaction:

$P \left(s\right) + \frac{5}{2} C {l}_{2} \left(g\right) \rightarrow P C {l}_{5} \left(s\right)$

Charge and mass are balanced as required. And we need (ii) equivalents weights of the reactants:

$\text{Moles of phosphorus}$ $=$ $\frac{0.567 \cdot g}{31.00 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.83 \times {10}^{-} 2 \cdot m o l$.

And thus we need $\frac{5}{2}$ equiv of chlorine gas; i.e. $\frac{5}{2} \times 1.83 \times {10}^{-} 2 \cdot m o l \times 70.90 \cdot g \cdot m o {l}^{-} 1$ $=$ ??g

Note that I treated phosphorus as $P$, when white phosphorus is actually ${P}_{4}$; I also knew that the elemental halogens are bimolecular. Capisce?