Aluminum oxide is formed when aluminum combines with oxygen in the air. How many grams of Al2O3 are formed when 13.6 mol of Al reacts?
4 Al + 3 O2 → 2 Al2O3
2/4 = 1/2 so half as many mols
13.6 / 2 = 6.8 mols of Al2O3
Al2O3 = 2*27+3*16 = 102 grams/mol
6.8 * 102 = ?
To find the number of grams of Al2O3 formed, we need to use the given molar ratio between Al and Al2O3.
From the balanced equation:
4 Al + 3 O2 → 2 Al2O3
We can see that 4 moles of Al react to form 2 moles of Al2O3. Therefore, the molar ratio is 4 moles Al : 2 moles Al2O3.
To find the number of moles of Al2O3 formed when 13.6 mol of Al reacts, we can set up a proportion:
4 moles Al / 2 moles Al2O3 = 13.6 mol Al / x moles Al2O3
Cross-multiplying, we get:
4 moles Al * x moles Al2O3 = 2 moles Al2O3 * 13.6 mol Al
Simplifying, we find:
4x moles Al2O3 = 27.2 mol Al2O3
Dividing both sides by 4, we get:
x moles Al2O3 = 27.2 mol Al2O3 / 4
x moles Al2O3 = 6.8 mol Al2O3
Finally, to find the grams of Al2O3 formed, we can use the molar mass of Al2O3, which is 101.96 g/mol.
grams Al2O3 = moles Al2O3 * molar mass Al2O3
grams Al2O3 = 6.8 mol Al2O3 * 101.96 g/mol
grams Al2O3 = 694.048 g
Therefore, when 13.6 mol of Al reacts, 694.048 grams of Al2O3 are formed.
To determine how many grams of Al2O3 are formed when 13.6 mol of Al reacts, we need to use the molar ratio between Al and Al2O3 from the balanced chemical equation:
4 Al + 3 O2 → 2 Al2O3
According to the equation, it takes 4 moles of Al to produce 2 moles of Al2O3.
Now, we can set up a proportion to solve for the grams of Al2O3:
(13.6 mol Al) x (2 mol Al2O3 / 4 mol Al) = x grams Al2O3
Simplifying the proportion:
(13.6 mol Al) x (1/2) = x grams Al2O3
x = 13.6 mol Al x (1/2)
x = 6.8 mol Al2O3
To convert moles to grams, we need to use the molar mass of Al2O3. Aluminum (Al) has a molar mass of 26.98 g/mol, and oxygen (O) has a molar mass of 16.00 g/mol.
Molar mass of Al2O3 = (2 x molar mass of Al) + (3 x molar mass of O)
Molar mass of Al2O3 = (2 x 26.98 g/mol) + (3 x 16.00 g/mol)
Molar mass of Al2O3 = 101.96 g/mol
Finally, we can calculate the mass of Al2O3 formed:
Mass of Al2O3 = 6.8 mol Al2O3 x 101.96 g/mol
Mass of Al2O3 = 691.808 g
Therefore, when 13.6 mol of Al reacts, approximately 691.808 grams of Al2O3 are formed.