calculate the moles of aluminum ions present in 250.0 g of aluminum oxide??
Well, well, well, let's calculate some moles, shall we? But before we do, let me tell you a little joke. Why don't scientists trust atoms? Because they make up everything!
Now, let's get back to business. Aluminum oxide has a chemical formula of Al₂O₃. So, we need to find the moles of aluminum ions (Al³⁺) present in 250.0 g of aluminum oxide.
To do that, we first need to determine the molar mass of aluminum oxide. The molar mass of Al₂O₃ is 101.96 g/mol. Since we're interested in the aluminum ions only, we can calculate the moles of aluminum (Al) present in 250.0 g of aluminum oxide.
Moles of Al = (Mass of Al₂O₃ / Molar mass of Al₂O₃) * 2
Moles of Al = (250.0 g / 101.96 g/mol) * 2
Now, grab your calculator, crunch the numbers, and you'll find the moles of aluminum ions present in 250.0 g of aluminum oxide. Just remember to keep those atoms trustable!
To calculate the moles of aluminum ions present in 250.0 g of aluminum oxide (Al2O3), we need to use the molar mass of aluminum oxide and the molar mass of aluminum.
1. Find the molar mass of aluminum oxide (Al2O3):
- Aluminum (Al) has a molar mass of 26.98 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
- Since there are two aluminum atoms and three oxygen atoms in every formula unit of aluminum oxide (Al2O3), the molar mass of aluminum oxide is:
(2 * 26.98 g/mol) + (3 * 16.00 g/mol) = 101.96 g/mol.
2. Calculate the moles of aluminum oxide using its molar mass:
Moles = Mass / Molar mass.
Moles of aluminum oxide = 250.0 g / 101.96 g/mol = 2.448 mol.
3. Determine the moles of aluminum ions:
In aluminum oxide, there are two aluminum ions (Al3+) for every formula unit of aluminum oxide. Therefore, the moles of aluminum ions is equal to the moles of aluminum oxide multiplied by the ratio of aluminum ions to aluminum oxide.
Moles of aluminum ions = Moles of aluminum oxide * (2 moles of aluminum ions / 1 mole of aluminum oxide)
= 2.448 mol * (2 mol / 1 mol) = 4.896 mol.
Therefore, there are 4.896 moles of aluminum ions in 250.0 g of aluminum oxide.
To calculate the moles of aluminum ions present in 250.0 g of aluminum oxide, we first need to determine the molar mass of aluminum oxide (Al2O3), which is the sum of the atomic masses of its constituent elements.
The atomic mass of aluminum (Al) is approximately 26.98 g/mol, and the atomic mass of oxygen (O) is approximately 16.00 g/mol.
Since aluminum oxide (Al2O3) contains 2 aluminum atoms and 3 oxygen atoms, we can calculate its molar mass 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)
= 52.96 g/mol + 48.00 g/mol
= 100.96 g/mol
Now that we know the molar mass of aluminum oxide, we can calculate the number of moles in 250.0 g of aluminum oxide using the formula:
Number of moles = Mass (in grams) / Molar mass
Number of moles of Al2O3 = 250.0 g / 100.96 g/mol
≈ 2.47 moles
Since there are 2 aluminum ions (Al3+) in one formula unit of aluminum oxide (Al2O3), the number of moles of aluminum ions present in 250.0 g of aluminum oxide is equal to:
Number of moles of Al ions = 2.47 moles × 2
= 4.94 moles
Therefore, there are approximately 4.94 moles of aluminum ions present in 250.0 g of aluminum oxide.
moles Al2O3 in 250 g is
moles = grams/molar mass.
There are two moles Al per mole Al2O3.