How do you identify the limiting reactant when 5.78 g of phosphorus react with 27.9 of liquid bromine?

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

1 Answer
Nov 28, 2016

Answer:

By taking the molar quantities of each reagent........

Explanation:

You have written the stoichiometric equation; we could use half-inegral coefficients to make the arithmetic a bit easier:

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

#"Moles of phosphorus"# #=# #(5.78)/(31.00*g*mol^-1)=0.181*mol#,

#"Moles of bromine"# #=# #(27.9)/(159.8*g*mol^-1)=0.175*mol#,

Clearly, the bromine is in stoichiometric deficiency. Why? Because stoichiometric equivalence would demand,

#0.181*molxx3/2xx159.8*g*mol~=44*g# bromine.

There are few things in the laboratory that are as nasty and corrosive as liquid bromine. It would cause horrendous burns, and it is treated with considerable respect.