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0.7g of iron reacts directly with 0.4g of sulphur to form ferrous sulphide. If 2.8g of iron dissolves in dil. HCL and excess of sodium sulphide solution is added, 4.4g of iron sulphide is precipitated. If it obeys the law of constant composition. What is the ratio of weights of Fe:S?
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Ratio of weights of Fe:S
To determine the ratio of weights of Fe:S, we need to analyze the given information and use the law of constant composition.
Law of Constant Composition:
The law of constant composition states that a pure compound always contains the same elements in the same proportions by mass. This means that regardless of the source or method of preparation, a particular compound will always have the same ratio of elements.
Given information:
1. 0.7g of iron reacts directly with 0.4g of sulphur to form ferrous sulphide.
2. 2.8g of iron dissolves in dil. HCL and excess of sodium sulphide solution is added, resulting in the precipitation of 4.4g of iron sulphide.
Let's analyze the given information step by step:
Step 1: Reaction between iron and sulfur
- According to the first statement, 0.7g of iron reacts with 0.4g of sulfur to form ferrous sulphide (FeS).
- From this information, we can determine the molar ratio between iron and sulfur in the reaction.
- The molar mass of iron (Fe) is 55.85 g/mol, and the molar mass of sulfur (S) is 32.07 g/mol.
- We can calculate the number of moles of iron and sulfur in the reaction using their respective masses and molar masses.
- Moles of Fe = 0.7g / 55.85 g/mol = 0.0125 mol
- Moles of S = 0.4g / 32.07 g/mol = 0.0125 mol
- The molar ratio of Fe:S in the reaction is 1:1.
Step 2: Reaction between iron and sodium sulfide
- In the second statement, 2.8g of iron dissolves in dil. HCL, and excess sodium sulfide solution is added.
- This results in the precipitation of 4.4g of iron sulfide (FeS).
- The formation of iron sulfide confirms the molar ratio of Fe:S as 1:1, as determined in the first step.
Conclusion:
Based on the given information and the law of constant composition, the ratio of weights of Fe:S is 1:1. This means that for every gram of iron (Fe) present, there will be an equal weight of sulfur (S) in the compound ferrous sulfide (FeS).
To determine the ratio of weights of Fe:S, we need to analyze the given information and use the law of constant composition.
Law of Constant Composition:
The law of constant composition states that a pure compound always contains the same elements in the same proportions by mass. This means that regardless of the source or method of preparation, a particular compound will always have the same ratio of elements.
Given information:
1. 0.7g of iron reacts directly with 0.4g of sulphur to form ferrous sulphide.
2. 2.8g of iron dissolves in dil. HCL and excess of sodium sulphide solution is added, resulting in the precipitation of 4.4g of iron sulphide.
Let's analyze the given information step by step:
Step 1: Reaction between iron and sulfur
- According to the first statement, 0.7g of iron reacts with 0.4g of sulfur to form ferrous sulphide (FeS).
- From this information, we can determine the molar ratio between iron and sulfur in the reaction.
- The molar mass of iron (Fe) is 55.85 g/mol, and the molar mass of sulfur (S) is 32.07 g/mol.
- We can calculate the number of moles of iron and sulfur in the reaction using their respective masses and molar masses.
- Moles of Fe = 0.7g / 55.85 g/mol = 0.0125 mol
- Moles of S = 0.4g / 32.07 g/mol = 0.0125 mol
- The molar ratio of Fe:S in the reaction is 1:1.
Step 2: Reaction between iron and sodium sulfide
- In the second statement, 2.8g of iron dissolves in dil. HCL, and excess sodium sulfide solution is added.
- This results in the precipitation of 4.4g of iron sulfide (FeS).
- The formation of iron sulfide confirms the molar ratio of Fe:S as 1:1, as determined in the first step.
Conclusion:
Based on the given information and the law of constant composition, the ratio of weights of Fe:S is 1:1. This means that for every gram of iron (Fe) present, there will be an equal weight of sulfur (S) in the compound ferrous sulfide (FeS).
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0.7g of iron reacts directly with 0.4g of sulphur to form ferrous sulphide. If 2.8g of iron dissolves in dil. HCL and excess of sodium sulphide solution is added, 4.4g of iron sulphide is precipitated. If it obeys the law of constant composition. What is the ratio of weights of Fe:S?
Question Description
0.7g of iron reacts directly with 0.4g of sulphur to form ferrous sulphide. If 2.8g of iron dissolves in dil. HCL and excess of sodium sulphide solution is added, 4.4g of iron sulphide is precipitated. If it obeys the law of constant composition. What is the ratio of weights of Fe:S? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 0.7g of iron reacts directly with 0.4g of sulphur to form ferrous sulphide. If 2.8g of iron dissolves in dil. HCL and excess of sodium sulphide solution is added, 4.4g of iron sulphide is precipitated. If it obeys the law of constant composition. What is the ratio of weights of Fe:S? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 0.7g of iron reacts directly with 0.4g of sulphur to form ferrous sulphide. If 2.8g of iron dissolves in dil. HCL and excess of sodium sulphide solution is added, 4.4g of iron sulphide is precipitated. If it obeys the law of constant composition. What is the ratio of weights of Fe:S?.
0.7g of iron reacts directly with 0.4g of sulphur to form ferrous sulphide. If 2.8g of iron dissolves in dil. HCL and excess of sodium sulphide solution is added, 4.4g of iron sulphide is precipitated. If it obeys the law of constant composition. What is the ratio of weights of Fe:S? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 0.7g of iron reacts directly with 0.4g of sulphur to form ferrous sulphide. If 2.8g of iron dissolves in dil. HCL and excess of sodium sulphide solution is added, 4.4g of iron sulphide is precipitated. If it obeys the law of constant composition. What is the ratio of weights of Fe:S? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 0.7g of iron reacts directly with 0.4g of sulphur to form ferrous sulphide. If 2.8g of iron dissolves in dil. HCL and excess of sodium sulphide solution is added, 4.4g of iron sulphide is precipitated. If it obeys the law of constant composition. What is the ratio of weights of Fe:S?.
Solutions for 0.7g of iron reacts directly with 0.4g of sulphur to form ferrous sulphide. If 2.8g of iron dissolves in dil. HCL and excess of sodium sulphide solution is added, 4.4g of iron sulphide is precipitated. If it obeys the law of constant composition. What is the ratio of weights of Fe:S? in English & in Hindi are available as part of our courses for NEET.
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