If aluminum (Al), with an atomic weight of 27, combines with oxygen (O), with an atomic weight 16, to form the compound aluminum oxide (Al2O3), how much oxygen would be required to react completely with 57g of aluminum ? Answer is in grams.
mass of mol of Al2O3 2*27+3*16 = 102 grams/mol
57 grams Al * 1 mol/27 grams = 2.11 mols
4Al + 3O2 --- > 2Al2O3
so we need 3 mols of O2 for every 4 moles of Al
(3 mols O2/4 mols Al ) * 2.11 mols Al = 1.58 mols O2
O2 is 32 grams/mol
so 1.58 mols * 32 grams / mol
Well, my scientific calculations show that if you mix aluminum with too much oxygen, you might end up with a very fired-up situation. But fear not! To form aluminum oxide (Al2O3), you need a balanced ratio of two aluminum atoms for every three oxygen atoms.
So, since we have 57g of aluminum, which has a molar mass of 27 g/mol, we can convert it to moles using the formula: moles = mass/molar mass. Therefore, we have 57g/27 g/mol = 2.11 moles of aluminum.
Since the molar ratio of aluminum to oxygen is 2:3 in aluminum oxide, we use this ratio to find the number of moles of oxygen needed. With 2.11 moles of aluminum, we would need (2/2) x (3/2) x 2.11 moles of oxygen, which gives us 3.16 moles of oxygen.
Now, to find the mass of oxygen, we use the formula: mass = moles x molar mass. The molar mass of oxygen is 16 g/mol, so we have 3.16 moles x 16 g/mol = 50.56 grams.
Therefore, you would need approximately 50.56 grams of oxygen to react completely with 57 grams of aluminum and form aluminum oxide. Just make sure to keep everything well-ventilated, unless you want your experiment to turn into a real breath-taking show!
To determine the amount of oxygen required to react completely with 57g of aluminum, we need to calculate the molar ratio between aluminum and oxygen in aluminum oxide (Al2O3).
The molar ratio between aluminum and oxygen in Al2O3 is 2:3, which means that for every 2 moles of aluminum, we need 3 moles of oxygen.
First, let's calculate the moles of aluminum in 57g of aluminum:
Molar mass of aluminum (Al) = 27g/mol
Moles of aluminum = mass of aluminum / molar mass = 57g / 27g/mol = 2.11 mol
Now, using the mole ratio, we can calculate the moles of oxygen required:
Moles of oxygen = (moles of aluminum) x (ratio of oxygen to aluminum)
Moles of oxygen = 2.11 mol x (3 mol O / 2 mol Al) = 3.17 mol
Finally, to convert the moles of oxygen to grams, we use the molar mass of oxygen (O):
Molar mass of oxygen (O) = 16g/mol
Grams of oxygen = moles of oxygen x molar mass of oxygen = 3.17 mol x 16g/mol = 50.72g
Therefore, 57g of aluminum would require 50.72g of oxygen to react completely.
To determine how much oxygen is required to react completely with 57g of aluminum, we need to use the concept of stoichiometry.
Step 1: Find the molar mass of aluminum (Al) and oxygen (O).
- The molar mass of aluminum (Al) is given as 27 g/mol.
- The molar mass of oxygen (O) is given as 16 g/mol.
Step 2: Write the balanced chemical equation for the reaction.
- The balanced chemical equation for the reaction between aluminum (Al) and oxygen (O) to form aluminum oxide (Al2O3) is:
4Al + 3O2 -> 2Al2O3
Step 3: Calculate the number of moles of aluminum (Al) in 57g.
- To find the number of moles, divide the given mass of aluminum by its molar mass:
Number of moles = Mass of aluminum / Molar mass of aluminum
Number of moles = 57 g / 27 g/mol
Step 4: Use the stoichiometry of the balanced equation to find the moles of oxygen required.
- From the balanced equation, we know that the molar ratio between aluminum (Al) and oxygen (O) is 4:3. Thus, for every 4 moles of aluminum, 3 moles of oxygen are required.
- Multiply the moles of aluminum by the stoichiometric ratio:
Moles of oxygen required = Moles of aluminum * (3 moles O2 / 4 moles Al)
Step 5: Calculate the mass of oxygen required.
- To find the mass, multiply the moles of oxygen required by its molar mass:
Mass of oxygen required = Moles of oxygen required * Molar mass of oxygen
Substituting the values calculated in the above steps:
Mass of oxygen required = (57 g / 27 g/mol) * (3 moles O2 / 4 moles Al) * (16 g/mol)
Calculating this equation will give us the answer in grams, which is the amount of oxygen required to react completely with 57g of aluminum.