Question

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?

Answer 1

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.