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A 12 g mixture of Na2CO3 and CaCO3 and clay when heated gave 1.792 L of CO2 and in another experiment same amount of mixture required 19.8 L of 0.01 N HCI for complete neutralization. Calculate weight of Na2CO3 present in sample.
Correct answer is '4'. Can you explain this answer?
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Verified Answer
A 12 g mixture of Na2CO3and CaCO3and clay when heated gave 1.792 L of ...
Na2CO3 does not get decomposed due to ignition. CaCO3 decomposes to form CaO and CO2
100 g of CaCO3 produced 22.4 L CO2
1.792 CO2 produced by = 100 x 1.792 / 22.4 = 8 g
100 g of CaCO3 produced 22.4 L CO2
1.792 CO2 produced by = 100 x 1.792 / 22.4 = 8 g
Most Upvoted Answer
A 12 g mixture of Na2CO3and CaCO3and clay when heated gave 1.792 L of ...
Understanding the Problem
To determine the weight of Na2CO3 in the mixture, we analyze the given data:
- Total weight of the mixture = 12 g
- CO2 evolved upon heating = 1.792 L
- Volume of HCl used for neutralization = 19.8 L of 0.01 N
Step 1: CO2 Evolution Analysis
- The reaction of Na2CO3 and CaCO3 upon heating releases CO2.
- Molar volume of gas at STP = 22.4 L/mol.
Calculating moles of CO2 produced:
- Moles of CO2 = Volume of CO2 / Molar volume = 1.792 L / 22.4 L/mol = 0.08 mol
The reaction stoichiometry:
- From CaCO3: 1 mol CaCO3 produces 1 mol CO2
- From Na2CO3: 1 mol Na2CO3 also produces 1 mol CO2
Hence, the total moles of CO2 = moles from Na2CO3 + moles from CaCO3.
Step 2: Neutralization Analysis
The moles of HCl used for neutralization:
- Moles of HCl = Volume (L) x Normality = 19.8 L x 0.01 N = 0.198 mol
Neutralization reactions:
- CaCO3 + 2 HCl → CaCl2 + H2O + CO2
- Na2CO3 + HCl → 2 NaCl + H2O + CO2
Considering stoichiometry, we can relate moles of Na2CO3 and CaCO3 to moles of HCl used.
Step 3: Calculate Weight of Na2CO3
Let x = moles of Na2CO3 and y = moles of CaCO3.
From CO2 evolution:
x + y = 0.08
From neutralization:
x + 2y = 0.198
Now, solving these equations:
1. From the first equation: y = 0.08 - x
2. Substituting into the second equation gives you x = 0.04 mol (Na2CO3) and y = 0.04 mol (CaCO3).
Weight of Na2CO3 = moles x molar mass = 0.04 mol x 106 g/mol = 4 g.
Conclusion
The weight of Na2CO3 present in the sample is 4 g.
To determine the weight of Na2CO3 in the mixture, we analyze the given data:
- Total weight of the mixture = 12 g
- CO2 evolved upon heating = 1.792 L
- Volume of HCl used for neutralization = 19.8 L of 0.01 N
Step 1: CO2 Evolution Analysis
- The reaction of Na2CO3 and CaCO3 upon heating releases CO2.
- Molar volume of gas at STP = 22.4 L/mol.
Calculating moles of CO2 produced:
- Moles of CO2 = Volume of CO2 / Molar volume = 1.792 L / 22.4 L/mol = 0.08 mol
The reaction stoichiometry:
- From CaCO3: 1 mol CaCO3 produces 1 mol CO2
- From Na2CO3: 1 mol Na2CO3 also produces 1 mol CO2
Hence, the total moles of CO2 = moles from Na2CO3 + moles from CaCO3.
Step 2: Neutralization Analysis
The moles of HCl used for neutralization:
- Moles of HCl = Volume (L) x Normality = 19.8 L x 0.01 N = 0.198 mol
Neutralization reactions:
- CaCO3 + 2 HCl → CaCl2 + H2O + CO2
- Na2CO3 + HCl → 2 NaCl + H2O + CO2
Considering stoichiometry, we can relate moles of Na2CO3 and CaCO3 to moles of HCl used.
Step 3: Calculate Weight of Na2CO3
Let x = moles of Na2CO3 and y = moles of CaCO3.
From CO2 evolution:
x + y = 0.08
From neutralization:
x + 2y = 0.198
Now, solving these equations:
1. From the first equation: y = 0.08 - x
2. Substituting into the second equation gives you x = 0.04 mol (Na2CO3) and y = 0.04 mol (CaCO3).
Weight of Na2CO3 = moles x molar mass = 0.04 mol x 106 g/mol = 4 g.
Conclusion
The weight of Na2CO3 present in the sample is 4 g.
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A 12 g mixture of Na2CO3and CaCO3and clay when heated gave 1.792 L of CO2and in another experiment same amount of mixture required 19.8 L of 0.01 N HCI for complete neutralization. Calculate weight of Na2CO3present in sample.Correct answer is '4'. Can you explain this answer?
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A 12 g mixture of Na2CO3and CaCO3and clay when heated gave 1.792 L of CO2and in another experiment same amount of mixture required 19.8 L of 0.01 N HCI for complete neutralization. Calculate weight of Na2CO3present in sample.Correct answer is '4'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 12 g mixture of Na2CO3and CaCO3and clay when heated gave 1.792 L of CO2and in another experiment same amount of mixture required 19.8 L of 0.01 N HCI for complete neutralization. Calculate weight of Na2CO3present in sample.Correct answer is '4'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 12 g mixture of Na2CO3and CaCO3and clay when heated gave 1.792 L of CO2and in another experiment same amount of mixture required 19.8 L of 0.01 N HCI for complete neutralization. Calculate weight of Na2CO3present in sample.Correct answer is '4'. Can you explain this answer?.
A 12 g mixture of Na2CO3and CaCO3and clay when heated gave 1.792 L of CO2and in another experiment same amount of mixture required 19.8 L of 0.01 N HCI for complete neutralization. Calculate weight of Na2CO3present in sample.Correct answer is '4'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 12 g mixture of Na2CO3and CaCO3and clay when heated gave 1.792 L of CO2and in another experiment same amount of mixture required 19.8 L of 0.01 N HCI for complete neutralization. Calculate weight of Na2CO3present in sample.Correct answer is '4'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 12 g mixture of Na2CO3and CaCO3and clay when heated gave 1.792 L of CO2and in another experiment same amount of mixture required 19.8 L of 0.01 N HCI for complete neutralization. Calculate weight of Na2CO3present in sample.Correct answer is '4'. Can you explain this answer?.
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A 12 g mixture of Na2CO3and CaCO3and clay when heated gave 1.792 L of CO2and in another experiment same amount of mixture required 19.8 L of 0.01 N HCI for complete neutralization. Calculate weight of Na2CO3present in sample.Correct answer is '4'. Can you explain this answer?, a detailed solution for A 12 g mixture of Na2CO3and CaCO3and clay when heated gave 1.792 L of CO2and in another experiment same amount of mixture required 19.8 L of 0.01 N HCI for complete neutralization. Calculate weight of Na2CO3present in sample.Correct answer is '4'. Can you explain this answer? has been provided alongside types of A 12 g mixture of Na2CO3and CaCO3and clay when heated gave 1.792 L of CO2and in another experiment same amount of mixture required 19.8 L of 0.01 N HCI for complete neutralization. Calculate weight of Na2CO3present in sample.Correct answer is '4'. Can you explain this answer? theory, EduRev gives you an
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