Question

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

Answer 1

Approx. `0.7*mol` with respect to `PCl_5`.

Explanation:

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

`1/4P_4(s) + 5/2Cl_2(g) rarr PCl_5(s)`

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

`P_4(s) + 10Cl_2(g) rarr 4PCl_5(s)`

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` `=` `(22.0*g)/(4xx31.00*g*mol^-1)=0.177*mol`

The chlorine gas oxidant is present in excess.

And thus, by the given stoichiometry with respect to phosphorus, we can make `0.177*molxx4=0.710*mol` with respect to `PCl_5`.

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