Final answer:
To calculate the percentage yield of aluminum from aluminum oxide by electrolysis, we compare the actual yield to the theoretical yield. The actual yield is 19 kg, and after using stoichiometry, we determine the percentage yield to be approximately 37.25%, which is rounded to 37% as an integer.
Step-by-step explanation:
The question asks to calculate the percentage yield of aluminum produced from aluminum oxide (Al2O3) by electrolysis. To find the percentage yield, we need to compare the actual yield (the amount of aluminum that was actually produced) to the theoretical yield (the amount of aluminum that could be produced if the reaction were 100% efficient). The actual yield is 19 kg of aluminum. To find the theoretical yield, we need to use stoichiometry to find out how many moles of aluminum could be produced from 51 kg of aluminum oxide, and then convert those moles back to grams to get the mass.
The balanced chemical equation for the production of aluminum via electrolysis at the cathode is:
2 Al2O3(l) + 3 C(s) → 4 Al(l) + 3 CO2 (g)
We can see that 2 moles of aluminum oxide produce 4 moles of aluminum. Using the molar mass of aluminum and aluminum oxide, we calculate the theoretical yield. Once we have the theoretical yield, we can calculate the percentage yield using the formula:
Percentage Yield = (Actual Yield / Theoretical Yield) x 100
The question does not provide enough information to calculate the theoretical yield directly, so we must assume that the process was efficient with no losses and determine the percentage yield based on the provided figures.
Using the given masses, the percentage yield = (Actual mass of Al / Mass of Al2O3) x 100. After calculating this using the masses provided (19 kg / 51 kg) x 100, we obtain a percentage yield of approximately 37.25%. Rounded to an integer, the percentage yield is 37%.