# Question Video: Calculating the Mass of Aluminum Produced from a Hall–Heroult Cell given the Current and Aluminum’s Molar Mass Chemistry • 10th Grade

The industrial extraction of aluminum using Hall–Héroult cells requires a very large electrical current of 120 kA. How much aluminum is produced per hour, taking the molar mass of aluminum to be 27.0 g/mol and one faraday of charge as 9.65 × 10^4 C? Give your answer in kilograms to 2 decimal places.

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### Video Transcript

The industrial extraction of aluminum using Hall–Héroult cells requires a very large electrical current of 120 kiloamps. How much aluminum is produced per hour, taking the molar mass of aluminum to be 27.0 grams per mole and one faraday of charge as 9.65 times 10 to the four coulombs? Give your answer in kilograms to two decimal places.

The extraction of aluminum is an electrolytic process. Aluminum ions in the molten phase are reduced to liquid aluminum metal. For every one mole of aluminum ions, three moles of electrons are needed, and this produces one mole of aluminum metal. We are asked how much aluminum is produced per hour. In other words, what mass of aluminum is produced?

We are given the molar mass of aluminum. We can see that we first need to calculate the number of moles of aluminum produced. And using this value and the molar mass, we can then determine the mass. We are not given the number of moles of aluminum, but we are given other data. We are given the current that flows in kiloamps and the time that has elapsed, which is one hour. We are also given the value of one faraday of charge.

A useful equation in electrolysis is 𝑄 equals 𝐼𝑡, where 𝑄 is the charge transferred in coulombs, 𝐼 the current in amperes, and 𝑡 the time in seconds. We need to convert the current to amperes and the time to seconds. Kiloamps can be converted to amps by multiplying by this conversion factor. Kiloamps cancel, and we get 120,000 amperes. Then, we can convert time in hours to minutes using this conversion factor and minutes to seconds using this conversion factor. Hours cancel, minutes cancel, and we get an answer of 3,600 seconds, which is equivalent to one hour.

Now we can work out the charge transferred. Using our key equation 𝑄 equals 𝐼𝑡, we can substitute in the current in amperes and the time in seconds. Solving, we get the charge transferred in one hour, which is 432,000,000 coulombs. The next step is to convert the amount of charge transferred to a number of moles of electrons transferred. There are various ways to do this. One way is to use the key equation 𝑄 equals 𝑛𝐹, where 𝑄 is the charge transferred in coulombs, 𝑛 the number of moles of electrons transferred, and 𝐹 the Faraday constant.

Now, we are given one faraday of charge as 9.65 times 10 to the four coulombs. The Faraday constant, however, is this same value of charge in coulombs but per one mole of electrons transferred. The Faraday constant is, therefore, the charge on one mole of elementary charge. So we can rearrange our key equation to give 𝑛 is equal to 𝑄 divided by 𝐹. Substituting the amount of charge that flowed in one hour divided by the charge on one mole of elementary charge, which is equivalent to the charge on one mole of electrons, we can cancel the coulomb units. We get the number of moles of charge transferred in an hour, which is 4,476.68 moles.

Let’s clear some space. The next step is to convert the moles of charge transferred to the moles of aluminum produced. From the balanced half equation, we know that three moles of charge produces one mole of aluminum. But we don’t have three moles of electrons or charge. We have 4,476.68 moles of charge. And this will produce 𝑥 moles of aluminum. We can then solve this ratio by taking the moles of charge that we have dividing it by three. And we can solve for the moles of aluminum produced, 1,492.23 moles, which is this value here.

Finally, we can use the key equation number of moles is equal to mass divided by molar mass to determine the mass of aluminum produced. We can rearrange the key equation and multiply moles by molar mass, substituting in the mole value that we have just calculated, 1,492.23 moles, multiplied by the given molar mass of 27.0 grams per mole. We calculate an answer of 40,290.21 grams of aluminum produced in one hour in this Hall–Héroult cell.

However, we were asked to give an answer in kilograms. So we can multiply the mass of aluminum in grams by this conversion factor. Grams cancel out, giving an answer in kilograms, which we must remember to round off to two decimal places, which is then 40.29 kilograms of aluminum produced in one hour.