How much bauxite (Al2O3) would you have to mine if you wanted to obtain 100.0 g Al metal from the ore?
To find out how much bauxite (Al2O3) you would need to mine to obtain 100.0 g of aluminum (Al) metal, we first need to determine the molar mass of Al2O3.
The molar mass of aluminum (Al) is approximately 26.98 g/mol, and the molar mass of oxygen (O) is about 16.00 g/mol. Since there are two atoms of aluminum in Al2O3, and three atoms of oxygen, we can calculate the molar mass of Al2O3 as follows:
Molar mass of Al2O3 = (2 × molar mass of Al) + (3 × molar mass of O)
= (2 × 26.98 g/mol) + (3 × 16.00 g/mol)
= 54.00 g/mol + 48.00 g/mol
= 102.00 g/mol
Now, using the molar mass of Al2O3, we can set up a proportion to find the amount of bauxite needed to obtain 100.0 g of Al metal:
(100.0 g / 1) = (x g / 102.00 g/mol)
To solve for x (the amount of bauxite needed), we can cross-multiply and then divide:
x = (100.0 g * 102.00 g/mol) / 1
= 10,200 g/mol
Therefore, you would need to mine approximately 10,200 grams (10.2 kg) of bauxite (Al2O3) to obtain 100.0 grams of aluminum (Al) metal.
4Al + 3O2 ==> 2Al2O3
mols Al you want = grams/atomic mass Al.
Using the coefficients in the balanced equation, convert mols Al to mols Al2O3.
Then grams Al2O3 = mols Al2O3 x molar mass Al2O3.
This assume bauxite is pure Al2O3 and it isn't. Actually this is not a good answer.