Question

In an experiment 2.4 g of iron oxide an
reduction with hydrogen yield 1.68g iron. .
in another experiment, 2.9g of iron oxide
give 2.03g of iron on reduction with
hydrogen show
that the above
data illustrate the law of constant proportion.​

Answer 1

Answer:

In experiment 1,

2.4 g of iron oxide gives 1.68 g iron on reduction with hydrogen.

FeO→Fe+O

2

Percent of Fe in FeO is =

2.4

1.68

×100=70

Therefore ratio of m

Fe

:m

O

::7:3

In experiment 2:

2.9 g of iron oxide gives 2.09 g iron on reduction.

Percent of Fe in FeO is =

2.9

2.09

×100=72

Therefore ratio of m

Fe

:m

O

≈7:3

When a given chemical compound contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation, it follows law of constant proportion.

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Answer 2

Answer:

In experiment 1,

2.4 g of iron oxide gives 1.68 g iron on reduction with hydrogen.

FeO→Fe+O

2

Percent of Fe in FeO is =

2.4

1.68

×100=70

Therefore ratio of m

Fe

:m

O

::7:3

In experiment 2:

2.9 g of iron oxide gives 2.09 g iron on reduction.

Percent of Fe in FeO is =

2.9

2.09

×100=72

Therefore ratio of m

Fe

:m

O

≈7:3

When a given chemical compound contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation, it follows law of constant proportion.

Explanation: