If 63.8 grams of aluminum metal (al) react with 72.3 grams of sulfur (s) in a synthesis reaction, how many grams of the excess reactant will be left over when the reaction is complete?

23.2 g of Al will be left over when the reaction is complete

Explanation:

2Al  +  3S  → Al₂S₃

1 mol of Al = 26.98 g

1 mol of S = 32.06 g

Mole = Mass / Molar mass

63.8 g/ 26.98 g/m = 2.36 mole of Al

72.3 g / 32.06 g/m = 2.25 mole of S

2 mole of Aluminun react with 3 mole of sulfur

2.36 mole of Al react with (2.36 .3)/2 = 3.54 m of S

As I have 2.25 mole of S, and I need 3.54 S, is my limiting reagent so the limiting in excess is the Al.

3 mole of S react with 2 mole of Al

2.25 mole of S react with (2.25 m . 2)/3 = 1.50 mole

I need 1.50 mole of Al and I have 2.36, that's why the Al is in excess.

2.36 mole of Al - 1.50 mole of Al = 0.86 mole

This is the quantity of Al without reaction.

Molar mass . mole = Mass →  26.98 g/m . 0.86 m = 23.2 g

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