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

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F e_{2} O_{3}

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F e_{3} O_{4}

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F e O_{2}

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Solution

The correct option is B $F e O$
Weight of Iron oxide $= 0.5434 g$
Weight of oxygen lost as $H_{2} O = 0.1210 g$
Therefore, the weight of Iron $= 0.5434 - 0.1210 = 0.4224 g$
Element
Amount
%
mole
Ratio
Iron
0.4224 g
77.73
1.39
1
Oxygen
0.1210 g
22.26
1.39
1
Thus, formula is $F e O$ .

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Similar questions

A certain oxide of iron contains 2.5 g of oxygen for every 7.0 g of iron. If it is regarded as mixture of FeO and $F e_{2} O_{3}$ in the weight ratio x : y, what is x : y, (atomic weight of iron = 56)

Q. A certain mixed oxide of iron contains

2.5 g

of oxygen for every

7 g

of iron. If it is regarded as a mixture of

F e O

and

F e_{2} O_{3}

in the weight ratio

a : b

, what is

a : b

?
(atomic weight of iron = 56)

Q. Calculate the weight of Iron which will be converted into its oxide by the action of 18 g of steam.
Given the reaction :

3 F e + 4 H_{2} O \to F e_{3} O_{4} + 4 H_{2}

(Atomic mass of

F e = 56 g

Calculate the percentage composition of iron in a ferric oxide (

F e_{2} O_{3}

(Atomic mass of

F e = 56

)

In the production of iron, intermediate are $F e O$ , $F e_{2} O_{3}$ , and $F e_{3} O_{4}$ , which are oxide . If true enter 1 else 0.

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Element	Amount	%	mole	Ratio
Iron	0.4224 g	77.73	1.39	1
Oxygen	0.1210 g	22.26	1.39	1

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) :

The correct option is B FeOWeight of Iron oxide =0.5434gWeight of oxygen lost as H2O=0.1210gTherefore, the weight of Iron =0.5434−0.1210=0.4224gElement Amount % mole RatioIron 0.4224 g 77.731.391Oxygen 0.1210 g22.261.391Thus, formula is FeO.

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 $(F e = 56)$ :

The correct option is B $F e O$
Weight of Iron oxide $= 0.5434 g$
Weight of oxygen lost as $H_{2} O = 0.1210 g$
Therefore, the weight of Iron $= 0.5434 - 0.1210 = 0.4224 g$
Element
Amount
%
mole
Ratio
Iron
0.4224 g
77.73
1.39
1
Oxygen
0.1210 g
22.26
1.39
1
Thus, formula is $F e O$ .