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Question

Answers

[A] Law of constant proportions.

[B] Law of multiple proportions.

[C] Law of reciprocal proportions.

[D] Law of conservation of mass.

Answer
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In the question, it is given to us that 2.4g of iron oxide on reduction with hydrogen gave 1.68g of iron in the first experiment. We can write the reaction as-

\[FeO\to Fe{{O}_{2}}\]

Here, the starting product was 2.4g and in the product there was 1.68g of iron.

Therefore, the percentage of iron in iron oxide is $\dfrac{1.68}{2.4}\times 100=70%$

As 70% of iron oxide is iron therefore, the remaining 30% will be oxygen.

Therefore we can write the ratio of mass of iron and the mass of oxygen in iron oxide in this experiment will be = 7:3

In the second experiment, 2.9g of iron oxide gave us 2.09g of iron on reduction.

Therefore, the percentage of iron in iron oxide in this experiment will be $\dfrac{2.09}{2.9}\times 100=72%$

Here, the percentage of iron is 72% therefore the remaining 28% will be oxygen.

Therefore, the ratio of mass of iron and oxygen in iron oxide in the second experiment will be = 72:28 $\simeq $7:3.

As we can see from the above calculation that the ratio of iron and oxygen in iron oxide is almost similar in both the experiments and the law of constant proportion states that a given chemical compound will always contain its components in fixed ratios irrespective of the method of it was prepared by or the source.

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