Question
In an experiment 4.68 g iron oxide on reduction with hydrogen yields 3.86 g of iron. In another experiment 3.88 g of iron oxide gives 3.2 g of iron on reduction with hydrogen. Prove that the above data illustrates the law of constant proportions.
Solution
In the first experiment:
Mass of iron oxide = Mass of iron oxide - Mass of iron = (4.68 - 3.86) = 0.82 g
Ratio of oxygen and iron = 0.82 : 3.86 = 1:4.707
In the second experiment:
The mass of iron oxide = 3.88 g
The mass of iron after reduction = 3.20 g
Mass of iron oxide = Mass of iron oxide - Mass of iron = (3.88-3.20) = 0.68g
Ratio of oxygen and iron=0.68:3.20 =1:4.706
Thus, the data illustrates the law of consant proportions as in both the experiments the ratio of oxygen and iron is the same.
Mass of iron oxide = Mass of iron oxide - Mass of iron = (4.68 - 3.86) = 0.82 g
Ratio of oxygen and iron = 0.82 : 3.86 = 1:4.707
In the second experiment:
The mass of iron oxide = 3.88 g
The mass of iron after reduction = 3.20 g
Mass of iron oxide = Mass of iron oxide - Mass of iron = (3.88-3.20) = 0.68g
Ratio of oxygen and iron=0.68:3.20 =1:4.706
Thus, the data illustrates the law of consant proportions as in both the experiments the ratio of oxygen and iron is the same.
