Question

The weight of oxygen required to completely burn with 27g of Al is?
(P.S. The answer is not 24)

Answer 1

Answer of 24 gms seems to be correct.  If some one says it not right, then he/she is probably wrong.


Reaction :

   4 Al + 3 O₂  ==>  2 Al₂ O₃

Atomic weight of Aluminum = 27
Atomic weight of Oxygen = 16

So 108 gms of Aluminum reacts with 96 grams of Oxygen to completely burn and become an oxide.

Hence, 27 grams of Aluminum reacts with 24 grams of Oxygen.


4 moles of Al  reacts with 3 moles of  Oxygen.  Or, One mole of Al reacts with  0.75 mole of Oxygen.

Answer 2

Answer:

48

Explanation:

4Al +3O2 gives 2Al2O3

for Al 54÷27 moles are formed(2)

by using stichomatry

moles of oxygen =3×2/4 =3/2

weight of oxygen(O2) =3×32/2=48gm