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

1 Answer
Mar 13, 2017

Answer:

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#