the major commercial aluminium ore is bauxite,AlO3.H2O(s).What is its molar mass?The analysis of a representive sample of bauxite shows it contains 70% by mass of Al2O3..What is the maximum mass of aluminium metal that could be obtained from 200kg of bauxite by the HAll electrolytic process?
Here is a sample problem I posted for stoichiometry. Use that as a guide. Most of these are worked about the same way. The starting mass will be 200,000 x 0.92 = ?? g Al2O3. I expect your book has the equation for the electrolysis of bauxite by the Hall process.
26.981+(3*16)+(4*1.008)+(2*16)
=111.013
asking solution
To determine the molar mass of bauxite, you need to calculate the molar mass of the different elements present in it.
1. Start by finding the atomic masses of each element in the formula: Al (aluminum), O (oxygen), and H (hydrogen).
- The atomic mass of Al is 26.98 g/mol.
- The atomic mass of O is 16.00 g/mol.
- The atomic mass of H is 1.01 g/mol.
2. Multiply the atomic masses by the respective subscripts in the formula and sum them up:
(1 x Al) + (3 x O) + (1 x H) = 26.98 + (3 x 16.00) + 1.01 = 102.99 g/mol
So, the molar mass of bauxite (AlO3.H2O) is approximately 102.99 g/mol.
To calculate the maximum mass of aluminum metal that could be obtained from 200 kg of bauxite, you need to consider the percentage of Al2O3 in bauxite. According to the given information, 70% of the mass of bauxite is Al2O3.
1. Calculate the mass of Al2O3 in 200 kg of bauxite:
Mass of Al2O3 = 70% x 200 kg = 0.70 x 200 kg = 140 kg
2. Convert the mass of Al2O3 to the mass of Al by multiplying by the molar mass ratio:
Mass of Al = Mass of Al2O3 x (2 x Atomic mass of Al / Formula mass of Al2O3)
The molar mass of Al2O3 is calculated as follows:
2 x Atomic mass of Al + 3 x Atomic mass of O = (2 x 26.98) + (3 x 16.00) = 101.96 g/mol
Mass of Al = 140 kg x (2 x 26.98 g / 101.96 g/mol) = 74.07 kg
Therefore, the maximum mass of aluminum metal that could be obtained from 200 kg of bauxite by the Hall electrolytic process is approximately 74.07 kg.