Aluminum oxide is formed by reaction of excess oxygen with aluminum.

4 Al(s) + 3 O2(g) -> 2 Al2O3(s)

Calculate the mass of aluminum required to form 18.2 kg of aluminum oxide if the reaction has a percent yield of 83.4%.

4Al + 3O2 ==> 2Al2O3

kilomols Al2O3 = kg/molar mass = kmols needed at 100%. Call that ?
kmol needed at 89.4% = ?.0.894
Convert kmols Al2O3 to kmols Al using the coefficients in the balanced equation.
Now convert kmols to grams. kmols x atomic mass = kg Al.

To calculate the mass of aluminum required to form 18.2 kg of aluminum oxide, we need to determine the amount of aluminum oxide produced in the reaction, taking into account the percent yield.

First, let's find the theoretical yield of aluminum oxide. The balanced equation tells us that 4 moles of aluminum react to form 2 moles of aluminum oxide. Therefore, the molar ratio between aluminum and aluminum oxide is 4:2 or 2:1.

To find the theoretical moles of aluminum oxide formed, we can use the stoichiometry:

Moles of aluminum oxide = 18.2 kg / molar mass of aluminum oxide

The molar mass of aluminum oxide (Al2O3) can be calculated by adding the atomic masses of aluminum (Al) and oxygen (O):

Molar mass of Al2O3 = (2 * atomic mass of Al) + (3 * atomic mass of O)

Next, we need to consider the percent yield of the reaction. The percent yield tells us what percentage of the theoretical yield is actually obtained. In this case, the percent yield is given as 83.4%.

To calculate the actual moles of aluminum oxide formed, we multiply the theoretical moles by the percent yield:

Actual moles of aluminum oxide = theoretical moles * percent yield

Finally, we can calculate the mass of aluminum required using the molar ratio between aluminum and aluminum oxide:

Mass of aluminum = actual moles of aluminum oxide * molar mass of aluminum * (2 moles of Al / 2 moles of Al2O3)

Remember to convert the mass to the desired unit (such as grams or kilograms).

Note: Make sure to use consistent units throughout the calculation to avoid errors.