Calculate the moles of aluminum ions present in 250.0 g aluminum oxide
To calculate the moles of aluminum ions present in aluminum oxide, we need to use the molar mass of aluminum oxide (Al2O3) and the given mass in grams.
1. Determine the molar mass of aluminum oxide (Al2O3):
- Atomic mass of Al = 26.98 u
- Atomic mass of O = 16.00 u
- Molar mass of Al2O3 = (2 * 26.98 u) + (3 * 16.00 u) = 101.96 u
2. Convert the given mass of aluminum oxide to moles:
- Mass = 250.0 g
- Moles = Mass / Molar mass
= 250.0 g / 101.96 g/mol
Therefore, the moles of aluminum oxide present in 250.0 g of aluminum oxide is:
Moles = 2.45 mol
3. Determine the moles of aluminum ions (Al3+) present:
- In each molecule of aluminum oxide (Al2O3), there are 2 moles of aluminum ions (Al3+).
- Therefore, the moles of aluminum ions (Al3+) = Moles of aluminum oxide × 2
= 2.45 mol × 2
Thus, the moles of aluminum ions present in 250.0 g of aluminum oxide is:
Moles of aluminum ions = 4.9 mol