A #5*g# mass of aluminum metal is oxidized by a #10*g# mass of sulfur. If a #12.5*g# mass of #"aluminum sulfide"# is isolated, what is the percentage yield?

1 Answer
anor277
Dec 30, 2017

Aluminum is the limiting reagent....

Explanation:

We assess the following redox reaction....

#2Al(s) + 3S(s) stackrel(Delta)rarrAl_2S_3 #

#"Moles of aluminum metal"=(5.00*g)/(27.0*g*mol^-1)=0.185*mol#

#"Moles of sulfur"=(10.00*g)/(32.06*g*mol^-1)=0.312*mol#

And thus the sulfur is in stoichiometric excess, and at most we can make #(0.185*mol)/2# with respect to #"aluminum sulfide"# #-=(0.185*mol)/2xx150.16*g*mol^-1=13.9*g#..

And thus #"% yield"=(12.5*g)/(13.9*g)xx100%=??%#

I leave it to you to calculate the mass of excess sulfur.

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