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On adding excess of CaCl2 to a solution containing Na2CO3 and NaHCO3 ,x g of precipitate is was obtained.On adding in drops to the filtrate,a further y g of precipitate is obtained .In another experiment to the same amount of solution in excess of CaCl2 was added ,boiled,and filtrated .The amount of precipitate in the second experiment would be a) x+ y b) x/2+ y/2 c) x +y/2 d) none of these answer is C .pls explain?
Most Upvoted Answer
On adding excess of CaCl2 to a solution containing Na2CO3 and NaHCO3 ,...
Solution:
Given:
- Excess of CaCl2 added to a solution containing Na2CO3 and NaHCO3 gives x g of precipitate.
- Adding drops of CaCl2 to the filtrate of the above step gives y g of precipitate.
- In another experiment, the same amount of solution in excess of CaCl2 was added, boiled, and filtrated.
To find: The amount of precipitate obtained in the second experiment.
Explanation:
1. Precipitation reaction between CaCl2 and Na2CO3/NaHCO3:
CaCl2 + Na2CO3 → CaCO3 + 2NaCl
CaCl2 + NaHCO3 → CaCO3 + NaCl + H2O + CO2
2. Calculation of the amount of precipitate in the first step:
The precipitate obtained in the first step is due to the reaction between the excess of CaCl2 and Na2CO3/NaHCO3 present in the solution.
Hence, x g of precipitate obtained in the first step is due to the reaction between CaCl2 and Na2CO3/NaHCO3.
3. Calculation of the amount of precipitate in the second step:
In the second step, the same amount of solution in excess of CaCl2 was added, boiled, and filtrated.
Since the excess of CaCl2 is already added in the first step, there is no CaCl2 left to react in the second step.
Thus, no precipitate would be obtained in the second step.
Therefore, the correct answer is option (d) none of these.
Note: The amount of precipitate obtained in the first step is x g. Adding drops of CaCl2 to the filtrate gives y g of precipitate. This means that not all of the Na2CO3/NaHCO3 has precipitated in the first step. On adding more CaCl2, the remaining Na2CO3/NaHCO3 precipitates, giving y g of precipitate.
Given:
- Excess of CaCl2 added to a solution containing Na2CO3 and NaHCO3 gives x g of precipitate.
- Adding drops of CaCl2 to the filtrate of the above step gives y g of precipitate.
- In another experiment, the same amount of solution in excess of CaCl2 was added, boiled, and filtrated.
To find: The amount of precipitate obtained in the second experiment.
Explanation:
1. Precipitation reaction between CaCl2 and Na2CO3/NaHCO3:
CaCl2 + Na2CO3 → CaCO3 + 2NaCl
CaCl2 + NaHCO3 → CaCO3 + NaCl + H2O + CO2
2. Calculation of the amount of precipitate in the first step:
The precipitate obtained in the first step is due to the reaction between the excess of CaCl2 and Na2CO3/NaHCO3 present in the solution.
Hence, x g of precipitate obtained in the first step is due to the reaction between CaCl2 and Na2CO3/NaHCO3.
3. Calculation of the amount of precipitate in the second step:
In the second step, the same amount of solution in excess of CaCl2 was added, boiled, and filtrated.
Since the excess of CaCl2 is already added in the first step, there is no CaCl2 left to react in the second step.
Thus, no precipitate would be obtained in the second step.
Therefore, the correct answer is option (d) none of these.
Note: The amount of precipitate obtained in the first step is x g. Adding drops of CaCl2 to the filtrate gives y g of precipitate. This means that not all of the Na2CO3/NaHCO3 has precipitated in the first step. On adding more CaCl2, the remaining Na2CO3/NaHCO3 precipitates, giving y g of precipitate.
Community Answer
On adding excess of CaCl2 to a solution containing Na2CO3 and NaHCO3 ,...
The ans to this ques would b (a)....am I right?
if not please tell the correct one.
if not please tell the correct one.
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On adding excess of CaCl2 to a solution containing Na2CO3 and NaHCO3 ,x g of precipitate is was obtained.On adding in drops to the filtrate,a further y g of precipitate is obtained .In another experiment to the same amount of solution in excess of CaCl2 was added ,boiled,and filtrated .The amount of precipitate in the second experiment would be a) x+ y b) x/2+ y/2 c) x +y/2 d) none of these answer is C .pls explain?
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On adding excess of CaCl2 to a solution containing Na2CO3 and NaHCO3 ,x g of precipitate is was obtained.On adding in drops to the filtrate,a further y g of precipitate is obtained .In another experiment to the same amount of solution in excess of CaCl2 was added ,boiled,and filtrated .The amount of precipitate in the second experiment would be a) x+ y b) x/2+ y/2 c) x +y/2 d) none of these answer is C .pls explain? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about On adding excess of CaCl2 to a solution containing Na2CO3 and NaHCO3 ,x g of precipitate is was obtained.On adding in drops to the filtrate,a further y g of precipitate is obtained .In another experiment to the same amount of solution in excess of CaCl2 was added ,boiled,and filtrated .The amount of precipitate in the second experiment would be a) x+ y b) x/2+ y/2 c) x +y/2 d) none of these answer is C .pls explain? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for On adding excess of CaCl2 to a solution containing Na2CO3 and NaHCO3 ,x g of precipitate is was obtained.On adding in drops to the filtrate,a further y g of precipitate is obtained .In another experiment to the same amount of solution in excess of CaCl2 was added ,boiled,and filtrated .The amount of precipitate in the second experiment would be a) x+ y b) x/2+ y/2 c) x +y/2 d) none of these answer is C .pls explain?.
On adding excess of CaCl2 to a solution containing Na2CO3 and NaHCO3 ,x g of precipitate is was obtained.On adding in drops to the filtrate,a further y g of precipitate is obtained .In another experiment to the same amount of solution in excess of CaCl2 was added ,boiled,and filtrated .The amount of precipitate in the second experiment would be a) x+ y b) x/2+ y/2 c) x +y/2 d) none of these answer is C .pls explain? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about On adding excess of CaCl2 to a solution containing Na2CO3 and NaHCO3 ,x g of precipitate is was obtained.On adding in drops to the filtrate,a further y g of precipitate is obtained .In another experiment to the same amount of solution in excess of CaCl2 was added ,boiled,and filtrated .The amount of precipitate in the second experiment would be a) x+ y b) x/2+ y/2 c) x +y/2 d) none of these answer is C .pls explain? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for On adding excess of CaCl2 to a solution containing Na2CO3 and NaHCO3 ,x g of precipitate is was obtained.On adding in drops to the filtrate,a further y g of precipitate is obtained .In another experiment to the same amount of solution in excess of CaCl2 was added ,boiled,and filtrated .The amount of precipitate in the second experiment would be a) x+ y b) x/2+ y/2 c) x +y/2 d) none of these answer is C .pls explain?.
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