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What is the mass of aluminum metal that can be deposited by passing a current of 5.0 A through an Al(NO3)3 solution for 45 minutes? (Assume 100% efficiency and a Faraday constant of 96,500 C/mol.)

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To calculate the mass of aluminum metal deposited, we can use the following steps:1. Calculate the total charge  in Coulombs  passed through the solution.2. Determine the number of moles of electrons involved in the reaction.3. Calculate the number of moles of aluminum deposited.4. Calculate the mass of aluminum deposited.Step 1: Calculate the total chargeTotal charge  Q  = Current  I   Time  t Q = 5.0 A   45 min  60 s/min  = 5.0 A  2700 s = 13,500 CStep 2: Determine the number of moles of electronsMoles of electrons  n  = Total charge  Q  / Faraday constant  F n = 13,500 C / 96,500 C/mol = 0.1399 molStep 3: Calculate the number of moles of aluminum depositedThe balanced half-reaction for the reduction of aluminum is:Al + 3e  AlFrom the stoichiometry of the reaction, 3 moles of electrons are required to deposit 1 mole of aluminum.Moles of Al = Moles of electrons / 3Moles of Al = 0.1399 mol / 3 = 0.0466 molStep 4: Calculate the mass of aluminum depositedMass of Al = Moles of Al  Molar mass of AlMass of Al = 0.0466 mol  26.98 g/mol = 1.257 gTherefore, the mass of aluminum metal that can be deposited by passing a current of 5.0 A through an Al NO3 3 solution for 45 minutes is approximately 1.257 g.

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