# How many moles of PCL_5 can be produced from 22.0 g of P_4 (and excess Cl_2)?

Mar 13, 2017

Approx. $0.7 \cdot m o l$ with respect to $P C {l}_{5}$.

#### Explanation:

We need (i) a stoichiometric equation to represent the reaction:

$\frac{1}{4} {P}_{4} \left(s\right) + \frac{5}{2} C {l}_{2} \left(g\right) \rightarrow P C {l}_{5} \left(s\right)$

If you want you can multiply out the coefficients to give integral numbers:

${P}_{4} \left(s\right) + 10 C {l}_{2} \left(g\right) \rightarrow 4 P C {l}_{5} \left(s\right)$

I find the multiplication a bit easier to deal with if you use the former equation.

And we need (ii), equivalent quantities of each reactant:

Moles of ${P}_{4}$ $=$ $\frac{22.0 \cdot g}{4 \times 31.00 \cdot g \cdot m o {l}^{-} 1} = 0.177 \cdot m o l$

The chlorine gas oxidant is present in excess.

And thus, by the given stoichiometry with respect to phosphorus, we can make $0.177 \cdot m o l \times 4 = 0.710 \cdot m o l$ with respect to $P C {l}_{5}$.

This represents a mass of 0.177*molxx4xx208.24*g*mol^-1=??g