The amount of oxygen involved in the reaction would be 20.7 g
The equation of rusting of iron is as follows:
[tex]4Fe + 3O_2 ---> 2Fe_2O_3[/tex]
The mole ratio of Fe to O2 is 4:3.
Mole of 48.3 g Fe = mass/molar mass
= 48.3/56
= 0.8625 moles
Equivalent mole of O2 = 0.8625 x 3/4
= 0.6469 moles
Mass of 0.6469 moles O2 = mole x molar mass
= 0.6469 x 32
= 20.7 g
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