Question

Elemental phosphorus reacts with chlorine gas according to the equation
P4(s)+6Cl2(g)→4PCl3(l)
A reaction mixture initially contains 45.39 g P4 and 130.4 g Cl2.

Once the reaction has reached completion, what mass (in g) of the excess reactant is left?

Answer 1

Hi,
To know the reagent in excess we must first start from who is the limiting reagent in the reaction (the one that ends first), therefore the other will be the excess reagent.

Reagents P and Cl, will be identified as limiting or in excess with the following relationship: by dividing the moles of each of these by their stoichiometric coefficient in the reaction, the lower value between them will be the limiting reagent.


MOLES
P
Established molecular weight 30,974 g / mol


 Cl
Established molecular weight 35,453 g / mol


Relationship to know the limiting reagent

P

(It is divided by 4, because the subindice of this is 4 and therefore indicates that they are 4 moles)



Cl


(There are 12, because the subindice of this is 2 and in front a number 6, which multiplied is 12 moles)


The lowest value obtained was that of the chlorine element, Cl, which the limiting reagent does, failing that the phosphorus, P, is excessively reactive.

Since we know the moles of chlorine needed in the reaction, we will start from that to know the moles of phosphorus (excess reagent) that will consume that amount of chlorine.

4 moles of phosphorus react with 12 moles of Cl

37.97g is the mass of the reactant that reacts, by less subtracting the initial mass of the phosphorus element, we can finally obtain the mass of the excess reagent that remains after the reaction:








7.416g is the remaining mass of the excess reagent.



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