Home
Class 12
CHEMISTRY
In an experiment, 2.4 g of iron oxide on...

In an experiment, 2.4 g of iron oxide on iron. In another experiment, 2.9 g of iron oxide gave 2.09 g of iron on reduction. Which law is ilustrated from the above data?

A

Law of constant proportions

B

Law of multiple proportions

C

Law of conservation of mass

D

Law of reciprocal proportions

Text Solution

AI Generated Solution

The correct Answer is:
To determine which law of chemistry is illustrated by the given data, we will analyze the mass of iron oxide and the mass of iron produced in two separate experiments. ### Step-by-Step Solution: 1. **Identify the Data from the Experiments**: - In the first experiment: - Mass of iron oxide = 2.4 g - Mass of iron produced = 1.68 g - In the second experiment: - Mass of iron oxide = 2.9 g - Mass of iron produced = 2.09 g 2. **Calculate the Mass of Oxygen in Each Experiment**: - For the first experiment: - Mass of oxygen = Mass of iron oxide - Mass of iron - Mass of oxygen = 2.4 g - 1.68 g = 0.72 g - For the second experiment: - Mass of oxygen = Mass of iron oxide - Mass of iron - Mass of oxygen = 2.9 g - 2.09 g = 0.81 g 3. **Calculate the Ratio of Masses of Iron to Oxygen**: - For the first experiment: - Ratio of iron to oxygen = Mass of iron / Mass of oxygen - Ratio = 1.68 g / 0.72 g = 2.33 (or simplified to 7:3) - For the second experiment: - Ratio of iron to oxygen = Mass of iron / Mass of oxygen - Ratio = 2.09 g / 0.81 g = 2.58 (or simplified to 7:3) 4. **Compare the Ratios**: - Both experiments yield the same ratio of iron to oxygen (7:3), indicating that the elements combine in a fixed proportion by mass. 5. **Conclusion**: - The law illustrated by the data is the **Law of Constant Proportions** (or Law of Definite Proportions), which states that a chemical compound always contains its component elements in fixed ratio by mass, regardless of the source or method of preparation.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • NTA JEE MOCK TEST 61

    NTA MOCK TESTS|Exercise CHEMISTRY|25 Videos
  • NTA JEE MOCK TEST 63

    NTA MOCK TESTS|Exercise CHEMISTRY|25 Videos

Similar Questions

Explore conceptually related problems

In an experiment 2.4g of iron oxide in reduction with hydrogen gave 1.68 g of iron. In another experimet, 2.7 g of iron oxide gave 1.89 g of iron on reduction. Which law is illustrated from the above data?

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

Knowledge Check

  • On reduction 1.644 g of hot iron oxide give 1.15g of iron. Evaluate the equivalent weight of iron.

    A
    18.62
    B
    19.13
    C
    18.95
    D
    12.95
  • Similar Questions

    Explore conceptually related problems

    In an experiment, 2.4 g of Iron oxide on reduction with Hydrogen yields 1.68 g of Iron. In another experiment, 2.9 g of Iron oxide give 2.03 g of Iron on reduction with Hydrogen. Show that the above data illustrates the law of constant proportion.

    In an experiment, 1.288g of copper oxide was obtained from 1.03 g of copper. In another experiment 3.672 g of copper oxide gave, on reduction, 2.938 g of copper. Show that these figures verify the law of constant proportions.

    In an experiment, 4.90 g of copper oxide was obtained from 3.92 g of copper. In another experiment, 4.55 g of copper oxide gave, on reduction, 3.64 g of copper. Show with the help of calculation that these figures verify the law of constant proportions.

    2.16 g of copper metal when treated with nitric acid followed by ignition of the nitrate gave 2.70 g of copper oxide. In another experiment, 1.15 g of copper oxide upon reduction with hydrogen gave 0.92 g of copper. Show that the above data illustrates the law of definite proportions.

    2.16 of copper metal when treated with nitric acid followed by ignition of the nitrate gave 2.70 g of copper oxide. In another experiment 1.15 g of copper oxide upon reduction with hydrogen gave 0.92 g of copper. Show that the above data illustrate the Law of Definite Proportions.

    Weight of copper oxide obtained by heating 2.16 g of metallic copper with HNO_(3) and subsequent ingnition was 2.70 g In another experient, 1.15 g of copper oxide on reduction yielded 0.92 g of copper. Show that the results illustrance the law of definite proportions.

    5.975 g of the higher oxide of metal gave 5.575 g lower oxide on heating. The quantity of the lower oxide gave 5.175 g of metal on reduction. Prove that these results are in accordance with the law of multiple proportion.