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 and the concept of stoichiometry.

First, we find the molar mass of aluminum oxide (Al2O3). The atomic mass of aluminum (Al) is 26.98 g/mol, and the atomic mass of oxygen (O) is 16.00 g/mol. Since aluminum oxide has two aluminum atoms, the molar mass of aluminum oxide is:

2(26.98 g/mol Al) + 3(16.00 g/mol O) = 101.96 g/mol Al2O3

Now we can set up the conversion factor to calculate the moles of aluminum ions:

1 mol Al2O3 / 101.96 g Al2O3

To find the moles, we need to know the mass of aluminum oxide. In this case, it is given as 250.0 g.

Now we can plug the values into the conversion factor:

250.0 g Al2O3 × (1 mol Al2O3 / 101.96 g Al2O3) = 2.448 mol Al2O3

Since aluminum ions are in a 1:2 ratio with aluminum oxide, there are double the moles of aluminum ions:

2.448 mol Al2O3 × 2 = 4.896 mol Al ions

Therefore, there are 4.896 moles of aluminum ions present in 250.0 g of aluminum oxide.

moles = grams/molar mass.