Question

An iron sample contains 18% Fe3O4. What is the amount so that it is precipitated as Fe2O3 which weights 0.4gram.

Answer 1

Answer:

here is ur ans

Explanation:

Fe3 O4 ⟶1.5Fe2 O3

1mol  ; 1.5mol

232g ; 240g

?        ; 0.40g

Fe3 O4  required by 0.40g Fe 2 O3  =  232 / 240  × 0.4  = 0.3867g(pure)  

                                           =   0.3867×  100/18  g

                                           =2.15g impure Fe3 O4 (18% pure).

Thank you

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