All the oxygen in a 0.5434 g sample of a pure oxide of iron is removed by reduction in a stream of H 2. The loss in weight is 0.1210 g. hence, formula of the iron oxide is ( Fe =56)

Question

All the oxygen in a 0.5434 g sample of a pure oxide of iron is removed by reduction in a stream of H 2. The loss in weight is 0.1210 g. hence, formula of the iron oxide is ( Fe =56) (A) FeO (B) Fe 2 O 3 (C) Fe 3 O 4 (D) FeO 2

Tardigrade Admin · Accepted Answer

Iron oxide =0.5434 g Oxygen lost as H 2 O =0.1210 g Iron =0.5434-0.1210=0.4224 g Element Amount % Mole Ratio Iron (56) 0 1224 g 77.73 1.39 1 Oxygen(16) 0.1210 g 22.26 1.39 1 Thus FeO

Q. All the oxygen in a $0.5434 g$ sample of a pure oxide of iron is removed by reduction in a stream of $H_{2}$ . The loss in weight is $0.1210 g$ . hence, formula of the iron oxide is $(F e = 56)$

$F e O$

$F e_{2} O_{3}$

$F e_{3} O_{4}$

$F e O_{2}$

Iron oxide $= 0.5434 g$
Oxygen lost as $H_{2} O = 0.1210 g$
Iron $= 0.5434 - 0.1210 = 0.4224 g$

Element Amount % Mole Ratio

Iron (56) 0 1224 g 77.73 1.39 1

Oxygen(16) 0.1210 g 22.26 1.39 1

Thus $F e O$

Q. All the oxygen in a 0.5434g sample of a pure oxide of iron is removed by reduction in a stream of H2​. The loss in weight is 0.1210g. hence, formula of the iron oxide is (Fe=56)

FeO

Fe2​O3​

Fe3​O4​

FeO2​