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Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0 × 104 amperes of current is passed through molten Al2O3 for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency, atomic mass Al = 27 g/mol)
- a)1.3 x 104 g
- b)2.4 x 105 g
- c)9.0 x 103 g
- d)8.1 x 104 g
Correct answer is option 'D'. Can you explain this answer?
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Understanding the Electrolysis of Al2O3
When aluminum oxide (Al2O3) is reduced through electrolysis, aluminum (Al) is produced at the cathode. The Faraday's laws of electrolysis state that the amount of substance produced is directly proportional to the total electric charge passed through the electrolyte.
Calculating Total Charge (Q)
To find the mass of aluminum produced, we first need to calculate the total electric charge (Q) using the formula:
Q = I × t
Where:
- I = current in amperes (A)
- t = time in seconds (s)
Given:
- I = 4.0 × 10^4 A
- t = 6 hours = 6 × 3600 seconds = 21600 s
Now, substituting the values:
Q = 4.0 × 10^4 A × 21600 s = 864 × 10^8 C
Calculating Moles of Aluminum
The electrochemical equivalent of aluminum can be calculated using Faraday's constant (F = 96500 C/mol). The reaction for aluminum production is:
\[ Al^{3+} + 3e^- \rightarrow Al \]
This shows that 3 moles of electrons produce 1 mole of Al. Therefore, the number of moles of aluminum produced (n) can be calculated as:
\[ n = \frac{Q}{3F} \]
Substituting the values:
n = \(\frac{864 \times 10^8 C}{3 \times 96500 C/mol}\) = 3000 moles of Al
Calculating Mass of Aluminum
Now, to find the mass (m) of aluminum produced, we use:
m = n × M
Where M is the molar mass of aluminum (27 g/mol):
m = 3000 moles × 27 g/mol = 81000 g or 8.1 × 10^4 g
Thus, the mass of aluminum produced is **8.1 × 10^4 g**.
Conclusion
The correct answer is option 'D'.
When aluminum oxide (Al2O3) is reduced through electrolysis, aluminum (Al) is produced at the cathode. The Faraday's laws of electrolysis state that the amount of substance produced is directly proportional to the total electric charge passed through the electrolyte.
Calculating Total Charge (Q)
To find the mass of aluminum produced, we first need to calculate the total electric charge (Q) using the formula:
Q = I × t
Where:
- I = current in amperes (A)
- t = time in seconds (s)
Given:
- I = 4.0 × 10^4 A
- t = 6 hours = 6 × 3600 seconds = 21600 s
Now, substituting the values:
Q = 4.0 × 10^4 A × 21600 s = 864 × 10^8 C
Calculating Moles of Aluminum
The electrochemical equivalent of aluminum can be calculated using Faraday's constant (F = 96500 C/mol). The reaction for aluminum production is:
\[ Al^{3+} + 3e^- \rightarrow Al \]
This shows that 3 moles of electrons produce 1 mole of Al. Therefore, the number of moles of aluminum produced (n) can be calculated as:
\[ n = \frac{Q}{3F} \]
Substituting the values:
n = \(\frac{864 \times 10^8 C}{3 \times 96500 C/mol}\) = 3000 moles of Al
Calculating Mass of Aluminum
Now, to find the mass (m) of aluminum produced, we use:
m = n × M
Where M is the molar mass of aluminum (27 g/mol):
m = 3000 moles × 27 g/mol = 81000 g or 8.1 × 10^4 g
Thus, the mass of aluminum produced is **8.1 × 10^4 g**.
Conclusion
The correct answer is option 'D'.
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Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0 × 104 amperes of current is passed through molten Al2O3 for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency, atomic mass Al = 27 g/mol)a)1.3 x 104gb)2.4x 105gc)9.0 x 103gd)8.1 x 104gCorrect answer is option 'D'. Can you explain this answer?
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Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0 × 104 amperes of current is passed through molten Al2O3 for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency, atomic mass Al = 27 g/mol)a)1.3 x 104gb)2.4x 105gc)9.0 x 103gd)8.1 x 104gCorrect answer is option 'D'. Can you explain this answer? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0 × 104 amperes of current is passed through molten Al2O3 for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency, atomic mass Al = 27 g/mol)a)1.3 x 104gb)2.4x 105gc)9.0 x 103gd)8.1 x 104gCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0 × 104 amperes of current is passed through molten Al2O3 for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency, atomic mass Al = 27 g/mol)a)1.3 x 104gb)2.4x 105gc)9.0 x 103gd)8.1 x 104gCorrect answer is option 'D'. Can you explain this answer?.
Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0 × 104 amperes of current is passed through molten Al2O3 for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency, atomic mass Al = 27 g/mol)a)1.3 x 104gb)2.4x 105gc)9.0 x 103gd)8.1 x 104gCorrect answer is option 'D'. Can you explain this answer? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0 × 104 amperes of current is passed through molten Al2O3 for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency, atomic mass Al = 27 g/mol)a)1.3 x 104gb)2.4x 105gc)9.0 x 103gd)8.1 x 104gCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0 × 104 amperes of current is passed through molten Al2O3 for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency, atomic mass Al = 27 g/mol)a)1.3 x 104gb)2.4x 105gc)9.0 x 103gd)8.1 x 104gCorrect answer is option 'D'. Can you explain this answer?.
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