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How many moles of electrons are required to produce 99.9 g of aluminum metal from a sample of molten aluminum fluoride?

Answer :

To determine the number of moles of electrons required to produce 99.9 g of aluminum metal from molten aluminum fluoride, we need to consider the stoichiometry of the reaction and the molar masses of the compounds involved. The answer is approximately 3.47 moles of electrons.

The reaction that occurs during the electrolysis of molten aluminum fluoride involves the reduction of aluminum ions (Al3+) to produce aluminum metal (Al) at the cathode. Each aluminum ion gains three electrons to form aluminum metal. The molar mass of aluminum is 26.98 g/mol, which means that 99.9 g of aluminum corresponds to approximately 3.7 moles.

To find the number of moles of electrons required, we can use the stoichiometry of the reaction. Since each aluminum ion requires three electrons, the number of moles of electrons needed is three times the number of moles of aluminum. Thus, 3.7 moles of aluminum would require 3.7 × 3 = 11.1 moles of electrons.

However, it's important to note that the molar mass of aluminum fluoride (AlF3) is necessary to determine the number of moles of aluminum. If the molar mass of aluminum fluoride is provided, we can calculate the number of moles of aluminum and, subsequently, the number of moles of electrons required using the stoichiometry of the reaction.

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