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How much electricity in term of Faraday is required to produce 40 gram of Al from Al2O3 ? (Atomic mass of Al = 27 g/mol?
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How much electricity in term of Faraday is required to produce 40 gram...
Al3++3e−→Al
1 mole (27 g) of Al will require 3 moles of electrons.
40 g of Al will require 273×40F=4.4 F
1 mole (27 g) of Al will require 3 moles of electrons.
40 g of Al will require 273×40F=4.4 F
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How much electricity in term of Faraday is required to produce 40 gram...
Calculation of Electricity Required to Produce 40g of Al from Al2O3
Step 1: Write the Balanced Equation
The first step is to write a balanced chemical equation for the reaction that takes place during the production of aluminium from aluminium oxide.
2Al2O3(s) → 4Al(l) + 3O2(g)
Step 2: Determine the Number of Moles of Al2O3
The next step is to determine the number of moles of aluminium oxide required to produce 40 grams of aluminium. We can do this by using the atomic mass of aluminium and the molar mass of aluminium oxide.
Molar mass of Al2O3 = (2 x 27) + (3 x 16) = 102 g/mol
Number of moles of Al2O3 = Mass of Al2O3 / Molar mass of Al2O3
Number of moles of Al2O3 = 40 / 102 = 0.3922 moles
Step 3: Determine the Number of Moles of Al
From the balanced chemical equation, we know that 4 moles of aluminium are produced for every 2 moles of aluminium oxide consumed.
Number of moles of Al = 4 x 0.3922 / 2 = 0.7844 moles
Step 4: Calculate the Quantity of Charge Required
The quantity of charge required to produce a certain amount of aluminium can be calculated using Faraday's laws of electrolysis. Faraday's laws state that the amount of product produced at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.
The quantity of charge required to produce one mole of aluminium can be calculated using the equation:
Q = n x F
where Q is the quantity of charge in coulombs, n is the number of moles of aluminium produced, and F is the Faraday constant, which is equal to 96,485 coulombs per mole of electrons.
Q = 0.7844 x 96,485 = 75,702 coulombs
Step 5: Convert Coulombs to Amperes and Time
The quantity of charge required to produce the aluminium can be converted into amperes and time using the equation:
Q = I x t
where I is the current in amperes and t is the time in seconds.
I = Q / t
t = Q / I
Assuming a current of 1 ampere, the time required to produce the aluminium can be calculated as follows:
t = 75,702 / 1 = 75,702 seconds
This is equivalent to:
t = 75,702 / 3600 = 21 hours and 1 minute
Step 6: Answer the Question
The quantity of electricity required to
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How much electricity in term of Faraday is required to produce 40 gram of Al from Al2O3 ? (Atomic mass of Al = 27 g/mol?
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How much electricity in term of Faraday is required to produce 40 gram of Al from Al2O3 ? (Atomic mass of Al = 27 g/mol? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about How much electricity in term of Faraday is required to produce 40 gram of Al from Al2O3 ? (Atomic mass of Al = 27 g/mol? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for How much electricity in term of Faraday is required to produce 40 gram of Al from Al2O3 ? (Atomic mass of Al = 27 g/mol?.
How much electricity in term of Faraday is required to produce 40 gram of Al from Al2O3 ? (Atomic mass of Al = 27 g/mol? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about How much electricity in term of Faraday is required to produce 40 gram of Al from Al2O3 ? (Atomic mass of Al = 27 g/mol? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for How much electricity in term of Faraday is required to produce 40 gram of Al from Al2O3 ? (Atomic mass of Al = 27 g/mol?.
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