Question

Amount of oxygen required for complete combustion of 27 g al is

Answer 1

Answer:

24 gram o2

Explanation:

4AL+3 O2------- 2Al2O3

4 mol need ---- 3 mole of O2

108-----96g

27-----96/108. * 27

24g answer

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Answer 2

Answer:

Al + O₂  ⇒  Al₃O₂

Now aluminum's mass number is 27

and 2 molecules of oxyen is 32  

So it will be needing 32gms of oxygen to burn.