Calculate the number of kilowatt-hours of electricity required to produce 1.0 * 103 kg (1 metric ton) of aluminum by electrolysis of Al3+ if the applied voltage is 4.50 V and the process is 45% efficient.

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welcome back everyone. How many kilowatt hours of electricity are needed to obtain 1.8 times 10 to the third power kilograms, One metric ton of chromium through the electrolysis of chromium three plus carry on. Given that this process has a 40% efficiency and the applied voltage is 5.10 volts. So if we want to get to chromium metal, we're going to have to undergo a three electron reduction of our chromium, catalon. Recall. That reduction means we gain electrons and this will correspond for each mole of our chromium metal, which will require three Faraday's. So next we want to call Faraday's constant which is equal to the value of 96,485 columns per mole. So assuming 100% efficiency of our electricity, We're going to begin with the info from the prompt that we have to obtain 1.0 times 10 to the third power kg of chromium. Where we're going to multiply to go from kilograms to grams by recalling that are prefix kilo tells us that we have 10 to the third power of our base unit grams. Now we're going to multiply by our next conversion factor to go from chromium and its molar mass from the periodic table, which we can see is the value of 51.9961 g for one mole of chromium. Next going from molds of chromium into Faraday's. We would recognize that for one mole of our chromium, we would require three Faraday's of energy. And because we need our final unit to be ca looms? we're going to multiply by another conversion factor where we would recall that we have equal to one Faraday. A total of 96,485 and sorry 996,485 columns. So canceling out our units, we can get rid of kilograms grams moles and Faraday's leaving us with columns and at 100% efficiency we're going to produce electricity of 5. times 10 to the ninth power columns. Now, based on 40% efficiency, We'll begin with our 100% efficiency calculation as 5.5701 times 10 to the 9th power columns. And we're going to divide this by 40% as a decimal. So 400.40 which is going to yield a result of 1.39 to 5 times 10 to the 10th power columns of energy that is required. Now, according to our prompt, we need our final unit to be in kilowatt hours. So we're going to recall that one volt is equal to one jewel per column and recall that one jewel is equivalent to or rather let's make a correction here. But the conversion factor that 3.6 times 10 to the six power jules is equivalent to one kilowatt hour is how we'll get to our final answer. So what we should have is our energy required 1.39 to 5 times 10 to the 10th power columns. Where we're going to convert from columns into jewels by recalling that from the prompt. Were given a voltage of 5.10 volts here. And so because we know that a vault is equal to one jewel per column, we would interpret that as a conversion factor that for one column we have 5.10 jewels. And so now we can cancel out columns and focus on going from jules two kilowatt hours by plugging in that conversion factor that we have for one kilowatt hour, 3.6 times 10 to the six power jewels. So now canceling out jewels were left with kilowatt hours as our final unit, which is going to yield our answer of two point oh times 10 to the fourth power kilowatt hours as our final answer for the amount of electricity needed to obtain our one metric ton of chromium through the electrolysis of chromium three plus carry on. So it's highlighted in yellow is our final answer. To complete this example. I hope everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video

Calculate the number of kilowatt-hours of electricity required to produce 1.0 * 103 kg (1 metric ton) of aluminum by electrolysis of Al3+ if the applied voltage is 4.50 V and the process is 45% efficient.

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Key Concepts

Electrolysis

Efficiency

Kilowatt-Hour (kWh)