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
anor277
Nov 28, 2016

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.

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