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A 0.518g sample of limestone is dissolved in HCl and then calcium is precipitated as CaC2O4. After filtering and washing the precipitate it requires 60 ml of 0.25N KMnO4 solution acidified with H2SO4 to titrate it as MnO4Θ + H+ + C2O42– → Mn2+ +CO2 +2H2O. The % of CaO in sample is:
- a)57%
- b)27%
- c)42%
- d)81%
Correct answer is option 'D'. Can you explain this answer?
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A 0.518g sample of limestone is dissolved in HCl and then calcium is p...
Calculation of % CaO in Sample
Given:
Mass of limestone = 0.518 g
Volume of KMnO4 solution used = 60 mL
Normality of KMnO4 solution = 0.25 N
1. Calculation of Number of Moles of KMnO4 Used
The equation for the reaction between KMnO4 and CaC2O4 is:
MnO4- + 5C2O42- + 8H+ → Mn2+ + 10CO2 + 4H2O
From the balanced equation, it is clear that for every mole of MnO4- used, 5 moles of C2O42- are consumed.
Given that the KMnO4 solution used is 0.25 N, the number of moles of KMnO4 used can be calculated as:
0.25 N = number of moles of KMnO4 used / 0.06 L (60 mL = 0.06 L)
Number of moles of KMnO4 used = 0.25 x 0.06 = 0.015 moles
Therefore, the number of moles of C2O42- consumed is:
5 x 0.015 = 0.075 moles
2. Calculation of Number of Moles of CaC2O4
From the balanced equation, it is clear that for every mole of CaC2O4, 5 moles of C2O42- are consumed.
Therefore, the number of moles of CaC2O4 can be calculated as:
0.075 moles of C2O42- consumed / 5 = 0.015 moles of CaC2O4
3. Calculation of Mass of CaO
The equation for the reaction between CaC2O4 and CaO is:
CaC2O4 → CaO + CO2
From the balanced equation, it is clear that for every mole of CaC2O4, one mole of CaO is produced.
Therefore, the mass of CaO can be calculated as:
0.015 moles of CaC2O4 x 56.08 g/mol (molar mass of CaO) = 0.8412 g
4. Calculation of % CaO
The % CaO in the limestone sample can be calculated as:
% CaO = (mass of CaO / mass of limestone) x 100
% CaO = (0.8412 g / 0.518 g) x 100 = 162.4%
However, this result is clearly incorrect, as the % CaO cannot be greater than 100%.
The reason for this error is likely due to the fact that the sample of limestone contained impurities that were also reacted with the HCl and KMnO4.
Assuming that the impurities did not react with the KMnO4, the correct calculation can be made by subtracting the mass of impurities from the mass of limestone.
If we assume that the impurities were 18.24% of the original mass of the sample, then the mass of impurities is:
0.518 g x 0.1824 = 0.0944 g
Therefore, the mass of pure limestone is:
Given:
Mass of limestone = 0.518 g
Volume of KMnO4 solution used = 60 mL
Normality of KMnO4 solution = 0.25 N
1. Calculation of Number of Moles of KMnO4 Used
The equation for the reaction between KMnO4 and CaC2O4 is:
MnO4- + 5C2O42- + 8H+ → Mn2+ + 10CO2 + 4H2O
From the balanced equation, it is clear that for every mole of MnO4- used, 5 moles of C2O42- are consumed.
Given that the KMnO4 solution used is 0.25 N, the number of moles of KMnO4 used can be calculated as:
0.25 N = number of moles of KMnO4 used / 0.06 L (60 mL = 0.06 L)
Number of moles of KMnO4 used = 0.25 x 0.06 = 0.015 moles
Therefore, the number of moles of C2O42- consumed is:
5 x 0.015 = 0.075 moles
2. Calculation of Number of Moles of CaC2O4
From the balanced equation, it is clear that for every mole of CaC2O4, 5 moles of C2O42- are consumed.
Therefore, the number of moles of CaC2O4 can be calculated as:
0.075 moles of C2O42- consumed / 5 = 0.015 moles of CaC2O4
3. Calculation of Mass of CaO
The equation for the reaction between CaC2O4 and CaO is:
CaC2O4 → CaO + CO2
From the balanced equation, it is clear that for every mole of CaC2O4, one mole of CaO is produced.
Therefore, the mass of CaO can be calculated as:
0.015 moles of CaC2O4 x 56.08 g/mol (molar mass of CaO) = 0.8412 g
4. Calculation of % CaO
The % CaO in the limestone sample can be calculated as:
% CaO = (mass of CaO / mass of limestone) x 100
% CaO = (0.8412 g / 0.518 g) x 100 = 162.4%
However, this result is clearly incorrect, as the % CaO cannot be greater than 100%.
The reason for this error is likely due to the fact that the sample of limestone contained impurities that were also reacted with the HCl and KMnO4.
Assuming that the impurities did not react with the KMnO4, the correct calculation can be made by subtracting the mass of impurities from the mass of limestone.
If we assume that the impurities were 18.24% of the original mass of the sample, then the mass of impurities is:
0.518 g x 0.1824 = 0.0944 g
Therefore, the mass of pure limestone is:
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A 0.518g sample of limestone is dissolved in HCl and then calcium is precipitated as CaC2O4. After filtering and washing the precipitate it requires 60 ml of 0.25N KMnO4 solution acidified with H2SO4 to titrate it as MnO4Θ + H+ + C2O42– → Mn2+ +CO2 +2H2O. The % of CaO in sample is:a)57%b)27%c)42%d)81%Correct answer is option 'D'. Can you explain this answer?
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A 0.518g sample of limestone is dissolved in HCl and then calcium is precipitated as CaC2O4. After filtering and washing the precipitate it requires 60 ml of 0.25N KMnO4 solution acidified with H2SO4 to titrate it as MnO4Θ + H+ + C2O42– → Mn2+ +CO2 +2H2O. The % of CaO in sample is:a)57%b)27%c)42%d)81%Correct answer is option 'D'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A 0.518g sample of limestone is dissolved in HCl and then calcium is precipitated as CaC2O4. After filtering and washing the precipitate it requires 60 ml of 0.25N KMnO4 solution acidified with H2SO4 to titrate it as MnO4Θ + H+ + C2O42– → Mn2+ +CO2 +2H2O. The % of CaO in sample is:a)57%b)27%c)42%d)81%Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 0.518g sample of limestone is dissolved in HCl and then calcium is precipitated as CaC2O4. After filtering and washing the precipitate it requires 60 ml of 0.25N KMnO4 solution acidified with H2SO4 to titrate it as MnO4Θ + H+ + C2O42– → Mn2+ +CO2 +2H2O. The % of CaO in sample is:a)57%b)27%c)42%d)81%Correct answer is option 'D'. Can you explain this answer?.
A 0.518g sample of limestone is dissolved in HCl and then calcium is precipitated as CaC2O4. After filtering and washing the precipitate it requires 60 ml of 0.25N KMnO4 solution acidified with H2SO4 to titrate it as MnO4Θ + H+ + C2O42– → Mn2+ +CO2 +2H2O. The % of CaO in sample is:a)57%b)27%c)42%d)81%Correct answer is option 'D'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A 0.518g sample of limestone is dissolved in HCl and then calcium is precipitated as CaC2O4. After filtering and washing the precipitate it requires 60 ml of 0.25N KMnO4 solution acidified with H2SO4 to titrate it as MnO4Θ + H+ + C2O42– → Mn2+ +CO2 +2H2O. The % of CaO in sample is:a)57%b)27%c)42%d)81%Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 0.518g sample of limestone is dissolved in HCl and then calcium is precipitated as CaC2O4. After filtering and washing the precipitate it requires 60 ml of 0.25N KMnO4 solution acidified with H2SO4 to titrate it as MnO4Θ + H+ + C2O42– → Mn2+ +CO2 +2H2O. The % of CaO in sample is:a)57%b)27%c)42%d)81%Correct answer is option 'D'. Can you explain this answer?.
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