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A 10 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 g. The % by mass of NaCl in the original mixture is
- a)15.2%
- b)67.9%
- c)21.8%
- d)32.1%
Correct answer is option 'B'. Can you explain this answer?
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A 10 g sample of a mixture of calcium chloride and sodium chloride is ...
Solution:
Given,
Mass of the mixture = 10 g
Mass of CaO obtained = 1.62 g
Let x be the % by mass of NaCl in the original mixture.
Then, the % by mass of CaCl2 in the original mixture = (100 - x)
When Na2CO3 is added to the mixture, it reacts with CaCl2 to form CaCO3, while NaCl remains unchanged.
The balanced chemical equation for the reaction is:
CaCl2 + Na2CO3 → CaCO3 + 2NaCl
From the equation, we can see that 1 mole of CaCl2 reacts with 1 mole of Na2CO3 to form 1 mole of CaCO3 and 2 moles of NaCl.
Therefore, the number of moles of CaCO3 formed from the given sample of CaCl2 can be calculated as follows:
Molar mass of CaCl2 = 111 g/mol
Number of moles of CaCl2 = Mass/Molar mass = (100 - x)/111
Number of moles of Na2CO3 required = Number of moles of CaCl2
Number of moles of CaCO3 formed = Number of moles of CaCl2
Molar mass of CaCO3 = 100 g/mol
Mass of CaCO3 formed = Number of moles of CaCO3 × Molar mass of CaCO3
= (100 - x)/111 × 100 g/mol
When CaCO3 is heated, it decomposes to form CaO and CO2.
The balanced chemical equation for the reaction is:
CaCO3 → CaO + CO2
From the equation, we can see that 1 mole of CaCO3 forms 1 mole of CaO.
Therefore, the number of moles of CaO obtained from the given sample of CaCO3 can be calculated as follows:
Molar mass of CaO = 56 g/mol
Number of moles of CaO obtained = Mass/Molar mass = 1.62/56
Now, equating the number of moles of CaO obtained from CaCO3 to the number of moles of CaCO3 formed from CaCl2, we get:
(100 - x)/111 × 100 g/mol = 1.62/56 mol
Solving this equation for x, we get:
x = 67.9%
Therefore, the % by mass of NaCl in the original mixture is 67.9%.
Given,
Mass of the mixture = 10 g
Mass of CaO obtained = 1.62 g
Let x be the % by mass of NaCl in the original mixture.
Then, the % by mass of CaCl2 in the original mixture = (100 - x)
When Na2CO3 is added to the mixture, it reacts with CaCl2 to form CaCO3, while NaCl remains unchanged.
The balanced chemical equation for the reaction is:
CaCl2 + Na2CO3 → CaCO3 + 2NaCl
From the equation, we can see that 1 mole of CaCl2 reacts with 1 mole of Na2CO3 to form 1 mole of CaCO3 and 2 moles of NaCl.
Therefore, the number of moles of CaCO3 formed from the given sample of CaCl2 can be calculated as follows:
Molar mass of CaCl2 = 111 g/mol
Number of moles of CaCl2 = Mass/Molar mass = (100 - x)/111
Number of moles of Na2CO3 required = Number of moles of CaCl2
Number of moles of CaCO3 formed = Number of moles of CaCl2
Molar mass of CaCO3 = 100 g/mol
Mass of CaCO3 formed = Number of moles of CaCO3 × Molar mass of CaCO3
= (100 - x)/111 × 100 g/mol
When CaCO3 is heated, it decomposes to form CaO and CO2.
The balanced chemical equation for the reaction is:
CaCO3 → CaO + CO2
From the equation, we can see that 1 mole of CaCO3 forms 1 mole of CaO.
Therefore, the number of moles of CaO obtained from the given sample of CaCO3 can be calculated as follows:
Molar mass of CaO = 56 g/mol
Number of moles of CaO obtained = Mass/Molar mass = 1.62/56
Now, equating the number of moles of CaO obtained from CaCO3 to the number of moles of CaCO3 formed from CaCl2, we get:
(100 - x)/111 × 100 g/mol = 1.62/56 mol
Solving this equation for x, we get:
x = 67.9%
Therefore, the % by mass of NaCl in the original mixture is 67.9%.
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A 10 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 g. The % by mass of NaCl in the original mixture isa)15.2%b)67.9%c)21.8%d)32.1%Correct answer is option 'B'. Can you explain this answer?
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A 10 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 g. The % by mass of NaCl in the original mixture isa)15.2%b)67.9%c)21.8%d)32.1%Correct answer is option 'B'. 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 10 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 g. The % by mass of NaCl in the original mixture isa)15.2%b)67.9%c)21.8%d)32.1%Correct answer is option 'B'. 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 10 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 g. The % by mass of NaCl in the original mixture isa)15.2%b)67.9%c)21.8%d)32.1%Correct answer is option 'B'. Can you explain this answer?.
A 10 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 g. The % by mass of NaCl in the original mixture isa)15.2%b)67.9%c)21.8%d)32.1%Correct answer is option 'B'. 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 10 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 g. The % by mass of NaCl in the original mixture isa)15.2%b)67.9%c)21.8%d)32.1%Correct answer is option 'B'. 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 10 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 g. The % by mass of NaCl in the original mixture isa)15.2%b)67.9%c)21.8%d)32.1%Correct answer is option 'B'. Can you explain this answer?.
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