4.8 g of sulfur and 5.4 g of aluminum react based on the
chemical equation below and 4.5 g of aluminum sulfide (Al2S3)
are recovered from the reaction, a small amount cannot be
recovered.
3S + 2A1 - Al2S3
Determine the percent yield of Al2S3. (Molar mass of S = 32.06
g/mol, molar mass of Al = 26.98 g/mol)

Explanation:

            3S       +       2Al =       Al₂S₃

        3 x 26.98       2 x 26.98        

         96.18gm        53.96 gm

sulfur is in short supply so sulphur will be the limiting reagent

4.8 gm sulphur will require 53.96 x 4.8 / 96.18 gm of Al

53.96 x 4.8 / 96.18 gm

= 2.69 gm

aluminium sulphide formed

=  4.8 + 2.69

= 7.49 gm

percentage of  yield = 4.5 x100 / 7.49

= 60 %

the answer is true. hope it .