# Question c6f97

Mar 24, 2017

You do realize that neither $P C {l}_{3}$ nor $\text{pockle3}$ are room temperature gases don't you? You could easily find the density of liquid $P O C {l}_{3}$, so your given volume is INCORRECT.

#### Explanation:

Perhaps, we should redo the problem using molar quantities:

PCl_3(l) + 1/2O_2(g) rarr ""^(-)O-P^(+)Cl_3(l)

$\text{Moles of phosphorus trichloride}$ $=$ $\frac{194 \cdot g}{137.33 \cdot g \cdot m o {l}^{-} 1}$

$= 1.41 \cdot m o l$

And thus we would get $\text{1.41 moles of phosphoryl chloride}$ $=$

=1.41*molxx153.33*g*mol^-1=??.

Phosphorus trichloride has a normal boiling point of $76.1$ ""^@C; pockle3 boils at $105.8$ ""^@C#. Of course, both liquids would have vapour pressures, but the question has been poorly proposed.

$\text{Pockle3}$ has a density of $1.65 \cdot g \cdot m {L}^{-} 1$

$\text{Volume}$ $\equiv$ $\text{Mass"/"Density}$

$\text{Volume}$ $=$ $\frac{216.2 \cdot g}{1.65 \cdot g \cdot m {L}^{-} 1} = 131 \cdot m L$.