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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