What is the mass of #PBr_3# that can be produced from this reaction?

The reaction between phosphorus and liquid bromine is outlined as: #2P(s) + 3Br_2(l) -> 2PBr_3(l)#

1 Answer
Dec 1, 2016

Answer:

You have not specified a starting mass of either reactant; let's say you start with #10*g# of phosphorus, and stoichiometric bromine.......

Explanation:

You have written the stoichiometric equation,

#P(s) + 3/2Br_2(l) rarr PBr_3(l)#

This equation unequivocally tells us that a #31*g# mass of phosphorus reacts with a #240*g# mass of bromine to give approx. #271*g# of #PBr_3#. From where did I get these masses?

In the problem proposed here, we started with #(10*g)/(31.0*g*mol^-1)=1/3*mol# phosphorus.

And thus, by the stoichiometry of the reaction, I need #1/3xx3/2=1/2*mol# bromine, i.e. #1/2*molxx160*g*mol=80*g# elemental bromine.

By the way, bromine is one of the most corrosive substances you can handle in the laboratory. It can cause horrendous burns that would (literally!) take months to heal. Elemental phosphorus is also quite reactive, but I would handle this in preference to bromine.