How many grams of Aluminum oxide will be produced when 2.44 grams of Aluminum burns in air (i.e. react with Oxygen)?
4Al + 3O2 ==> 2Al2O3
1. mols Al - grams/atomic mass
2. Using the coefficients in the balanced equation, convert mols Al to mols Al2O3
3. Now convert mols Al2O3 to g. g = mols x molar mass
To determine the grams of aluminum oxide produced when aluminum reacts with oxygen, we need to understand the balanced chemical equation for this reaction.
The balanced chemical equation is:
4Al (s) + 3O2 (g) → 2Al2O3 (s)
From the equation, we can see that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide.
To find the moles of aluminum oxide produced, we can use the molar mass of aluminum (26.98 g/mol) and the molar mass of aluminum oxide (101.96 g/mol).
First, we need to calculate the number of moles of aluminum using its molar mass:
moles of aluminum = mass of aluminum / molar mass of aluminum
= 2.44 g / 26.98 g/mol
moles of aluminum = 0.0904 mol
From the balanced equation, we know that 4 moles of aluminum react with 2 moles of aluminum oxide. Therefore, we can determine the moles of aluminum oxide produced:
moles of aluminum oxide = (moles of aluminum / 4) × 2
= (0.0904 mol / 4) × 2
moles of aluminum oxide = 0.0452 mol
Finally, let's calculate the mass of aluminum oxide using its molar mass:
mass of aluminum oxide = moles of aluminum oxide × molar mass of aluminum oxide
= 0.0452 mol × 101.96 g/mol
mass of aluminum oxide = 4.61 grams
Therefore, approximately 4.61 grams of aluminum oxide will be produced when 2.44 grams of aluminum reacts with oxygen.