Aluminum cookware is reffered ti as "anodized aluminum". The anodizing process puts a layer of aluminum oxide on the aluminum, protecting it from corrosion. Consider the reaction of 10.00 grams of aluminum with 10.00 grams of oxygen to form aluminum oxide. What is the formula of aluminum oxide? Write a balance chemical equation for the reaction. Determine which reactant is the limiting reagent. How many grams of aluminum oxide will form? How many grams of excess reagent are left?
THANK YOU
The reaction for the formation of aluminum oxide is:
4Al + 3O2 -->2 Al2O3
Moles of Al = 10.00g Al / 26.98 g/mol = 0.3706 mol
Moles of O2 = 10.00 g O2 / 31.999 g/mol = 0.3125 mol O2.
(mol. Al)/(mol.O2) = 0.3706 / 0.3125 = 1.186 (based on amounts used)
(mol. Al)/(mol.O2) = 4/3 = 1.333 (based on the chemical equation)
Since the amount of Al needed is LESS that the amount available, aluminum is the LIMITING REAGENT.
The calculation of the amount of Al2O3 MUST be based on the limiting reagent. Al. Now you set up:
(0.3706 mol Al)(2 mol Al2O3 / 4 mol Al) = 0.1853 moles Al2O3.
Last step: Convert moles to grams.
Sorry for a bad sentence in my answer:
"Since the amount of Al needed is LESS that the amount available, aluminum is the LIMITING REAGENT. ".
It should read:
Since the amount of Al needed is MORE that the amount available, aluminum is the LIMITING REAGENT.
To determine the formula of aluminum oxide, we need to consider the valency of aluminum and oxygen. The valency of aluminum is +3, while the valency of oxygen is -2. In order to balance the charges between aluminum and oxygen, we need two aluminum ions (3+ × 2 = 6+) and three oxide ions (2- × 3 = 6-). Therefore, the formula for aluminum oxide is Al₂O₃.
Now let's write a balanced chemical equation for the reaction:
4Al + 3O₂ → 2Al₂O₃
To determine the limiting reagent, we compare the number of moles of each reactant to the stoichiometric coefficients in the balanced equation. First, we calculate the number of moles of aluminum and oxygen:
Number of moles of aluminum = mass of aluminum (g) / molar mass of aluminum
= 10.00 g / 27.0 g/mol (molar mass of aluminum)
≈ 0.370 moles
Number of moles of oxygen = mass of oxygen (g) / molar mass of oxygen
= 10.00 g / 32.0 g/mol (molar mass of oxygen)
≈ 0.3125 moles
From the balanced equation, we can see that the stoichiometric ratio between aluminum and oxygen is 4:3. Therefore, we need 0.3125 × (4/3) ≈ 0.4167 moles of aluminum for complete reaction. Since we have only 0.370 moles of aluminum, it is the limiting reagent.
Now, let's calculate the number of grams of aluminum oxide formed:
Number of moles of aluminum oxide = (0.370 moles of aluminum) × (2 moles of Al₂O₃ / 4 moles of Al)
= 0.370 × 2/4
≈ 0.185 moles
Mass of aluminum oxide formed = (number of moles of aluminum oxide) × (molar mass of aluminum oxide)
= 0.185 moles × (102.0 g/mol)
≈ 18.9 grams
To calculate the grams of excess reagent left, we need to determine the moles of oxygen that react with the limiting reagent:
Moles of oxygen reacted = (0.370 moles of aluminum) × (3 moles of O₂ / 4 moles of Al)
= 0.370 × 3/4
≈ 0.2775 moles
Moles of excess oxygen = (number of moles of oxygen initially) - (moles of oxygen reacted)
= 0.3125 moles - 0.2775 moles
≈ 0.035 moles
Mass of excess oxygen left = (moles of excess oxygen) × (molar mass of oxygen)
= 0.035 moles × (32.0 g/mol)
≈ 1.12 grams
Therefore, after the reaction, approximately 18.9 grams of aluminum oxide will be formed, and about 1.12 grams of oxygen will be left as an excess reagent.
To determine the formula of aluminum oxide, we need to balance the chemical equation for the reaction between aluminum and oxygen:
2 Al + 3 O2 -> 2 Al2O3
This equation shows that 2 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide.
To find the limiting reagent, we need to compare the amounts of each reactant.
First, let's convert the given masses of aluminum and oxygen into moles:
For aluminum:
10.00 g Al * (1 mole Al / atomic mass of Al) = 0.353 moles Al
For oxygen:
10.00 g O2 * (1 mole O2 / molecular mass of O2) = 0.313 moles O2
Now, we can calculate the moles of aluminum that react with the moles of oxygen:
Aluminum: 0.353 moles Al * (2 moles Al2O3 / 2 moles Al) = 0.353 moles Al2O3
Oxygen: 0.313 moles O2 * (2 moles Al2O3 / 3 moles O2) = 0.209 moles Al2O3
From the calculations, we can see that the oxygen is the limiting reagent because it produces fewer moles of Al2O3 compared to the aluminum.
Next, let's calculate the grams of aluminum oxide formed:
Using the balanced equation, we know that 2 moles of Al2O3 are produced from 2 moles of aluminum.
The molar mass of Al2O3 is the sum of the molar masses of aluminum and oxygen:
Al2O3 = (2 * atomic mass of Al) + (3 * atomic mass of O) = (2 * 26.98 g/mol) + (3 * 16.00 g/mol) = 101.96 g/mol
Now, let's calculate the grams of Al2O3 formed:
0.353 moles Al2O3 * (101.96 g Al2O3 / 1 mole Al2O3) = 36.00 g Al2O3
Therefore, 36.00 grams of aluminum oxide will form.
Lastly, let's determine the grams of excess reagent left. We can do this by calculating the moles of the excess reagent and then converting it to grams:
Oxygen: 0.313 moles O2 - 0.209 moles Al2O3 = 0.104 moles O2
0.104 moles O2 * (32.00 g O2 / 1 mole O2) = 3.33 g O2
Therefore, 3.33 grams of excess oxygen will be left unreacted.