How many grams of aluminum oxide can be formed by the reaction of 41.2 g of aluminum with oxygen?
4Al + 3O2 ==> 2Al2O3
mols Al = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Al to mols Al2O3.
Finally, convert mols Al2O3 to grams. g = mols x molar mass = ?
To determine the amount of aluminum oxide formed, we need to use the balanced chemical equation for the reaction between aluminum and oxygen.
2Al + 3O₂ → 2Al₂O₃
From the balanced equation, we can see that 2 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide.
To find the amount of aluminum oxide formed, we need to convert the mass of aluminum (41.2 g) to the moles of aluminum, then use the mole ratio to determine the moles of aluminum oxide, and finally convert the moles of aluminum oxide to grams.
Step 1: Convert the mass of aluminum to moles.
To convert grams to moles, we need to know the molar mass of aluminum, which is 26.98 g/mol.
Moles of aluminum = mass of aluminum (g) / molar mass of aluminum (g/mol)
Moles of aluminum = 41.2 g / 26.98 g/mol
Moles of aluminum = 1.53 mol
Step 2: Use the mole ratio to determine the moles of aluminum oxide.
From the balanced equation, we know that 2 moles of aluminum react to produce 2 moles of aluminum oxide.
Moles of aluminum oxide = Moles of aluminum * (2 moles of aluminum oxide / 2 moles of aluminum)
Moles of aluminum oxide = 1.53 mol * (2/2)
Moles of aluminum oxide = 1.53 mol
Step 3: Convert moles of aluminum oxide to grams.
To convert moles to grams, we need to know the molar mass of aluminum oxide, which is 101.96 g/mol.
Mass of aluminum oxide = moles of aluminum oxide * molar mass of aluminum oxide (g/mol)
Mass of aluminum oxide = 1.53 mol * 101.96 g/mol
Mass of aluminum oxide = 155.96 g
Therefore, 155.96 grams of aluminum oxide can be formed by the reaction of 41.2 grams of aluminum with oxygen.