Aluminum metal is exposed to atmospheric oxygen, it is oxidized to form aluminum oxide. How much heat is released by the complete oxidation of 24.2 grams of aluminium at 25°C and 1atm.
You need to look up the heat of formation of Al2O3 (Hof). I'm sure you have a set of tables in your text in kJ/mol.
Then ?kJ/mol x # mol in the equation x (24.2/atomic mass Al) = ?
To calculate the amount of heat released by the complete oxidation of aluminum, we first need to determine the balanced equation for the reaction. The oxidation of aluminum can be represented by the equation:
4 Al (s) + 3 O₂ (g) → 2 Al₂O₃ (s)
From the balanced equation, we can see that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide.
1 mole of aluminum has a molar mass of 26.98 g/mol, and 1 mole of aluminum oxide has a molar mass of 101.96 g/mol.
To calculate the amount of heat released, we will use the equation:
q = m × ΔH
Where:
q = heat released (in Joules)
m = mass of aluminum (in grams)
ΔH = enthalpy change (in J/g)
First, let's calculate the moles of aluminum:
moles of aluminum = mass of aluminum / molar mass of aluminum
moles of aluminum = 24.2 g / 26.98 g/mol
moles of aluminum = 0.897 mol
Next, let's calculate the moles of aluminum oxide formed:
moles of aluminum oxide = (2/4) × moles of aluminum
moles of aluminum oxide = (2/4) × 0.897 mol
moles of aluminum oxide = 0.4485 mol
Now, let's calculate the heat released:
ΔH = ΔH formation of Al₂O₃ = -1676 kJ/mol
q = moles of aluminum oxide × ΔH
q = 0.4485 mol × (-1676 kJ/mol)
q = -751.14 kJ
Finally, let's convert the heat from kilojoules to joules:
q (in Joules) = -751.14 kJ × 1000 J/kJ
q (in Joules) = -751,140 J
Therefore, the complete oxidation of 24.2 grams of aluminum at 25°C and 1 atm releases approximately -751,140 Joules of heat. The negative sign indicates that the reaction is exothermic, releasing heat.
To calculate the heat released by the complete oxidation of aluminum, we need to use the concept of enthalpy changes (ΔH) and stoichiometry.
The balanced equation for the oxidation of aluminum is:
4 Al + 3 O2 -> 2 Al2O3
According to this equation, 4 moles of aluminum react with 3 moles of oxygen gas to produce 2 moles of aluminum oxide.
1. First, calculate the moles of aluminum in 24.2 grams using the molar mass of aluminum.
Molar mass of aluminum (Al) = 26.98 g/mol
Moles of aluminum = mass / molar mass
Moles of aluminum = 24.2 g / 26.98 g/mol
2. Next, use the stoichiometry of the balanced equation to find the moles of aluminum oxide produced.
From the balanced equation, we know that:
4 moles of Al produces 2 moles of Al2O3
So, moles of Al2O3 = (moles of Al / 4) * 2
3. Calculate the heat released using the heat of formation (ΔHf) values.
The heat of formation of Al2O3 is -1669.8 kJ/mol.
Heat released = moles of Al2O3 * ΔHf
Now we have all the information to calculate the heat released.
Please note that we are assuming the reaction occurs under constant pressure conditions (1 atm) i.e., the enthalpy change is equal to the heat released.
Let's solve the calculations:
1. Moles of aluminum:
Moles of aluminum = 24.2 g / 26.98 g/mol = 0.897 mol (rounded to three decimal places)
2. Moles of aluminum oxide:
Moles of Al2O3 = (0.897 mol / 4) * 2 = 0.449 mol
3. Heat released:
Heat released = 0.449 mol * (-1669.8 kJ/mol)
Heat released ≈ -748.55 kJ (rounded to two decimal places)
So, the heat released by the complete oxidation of 24.2 grams of aluminum at 25°C and 1 atm is approximately -748.55 kJ.