What mass of chlorine gas is required to react with a #0.567*g# mass of phosphorus to make #PCl_5#?

1 Answer
anor277
Sep 26, 2016

A bit over #3*g# of #Cl_2# gas.

Explanation:

We need (i) a chemical reaction:

#P(s) + 5/2Cl_2(g) rarr PCl_5(s)#

Charge and mass are balanced as required. And we need (ii) equivalents weights of the reactants:

#"Moles of phosphorus"# #=# #(0.567*g)/(31.00*g*mol^-1)# #=# #1.83xx10^-2*mol#.

And thus we need #5/2# equiv of chlorine gas; i.e. #5/2xx1.83xx10^-2*molxx70.90*g*mol^-1# #=# #??g#

Note that I treated phosphorus as #P#, when white phosphorus is actually #P_4#; I also knew that the elemental halogens are bimolecular. Capisce?

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