Given: 2Al2O3 (s) 4Al (s) + 3O2 (g) °H = 3351.4 kJ, what is °H for the formation of 12.50 g of aluminum oxide?
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The deltaH for the decomposition of 1 mole of Al2O3 is half of 351.4 kJ since the chemical equation shows 2 moles of Al2O3.
Convert 12.50g Al2O3 to moles by dividing it by the molar mass of Al2O3. Multiply your answer by 351.4kJ to get the enthalpy of decomposition of 12.50g Al2O3.
The deltaH of formation is the inverse of the enthalpy of decompositin. Attach a negative sign to the previous answer.
To find °H for the formation of 12.50 g of aluminum oxide (Al2O3), we need to use the molar mass of Al2O3 and apply stoichiometry.
Step 1: Find the molar mass of Al2O3
The molar mass of Al2O3 can be calculated by summing the atomic masses of its constituent elements:
2(Al) + 3(O) = 2(26.98 g/mol) + 3(16.00 g/mol) = 101.96 g/mol
Step 2: Convert the given mass of aluminum oxide to moles
To convert grams to moles, divide the given mass by the molar mass:
12.50 g / 101.96 g/mol = 0.1224 mol
Step 3: Apply stoichiometry to determine the moles of Al required
The balanced chemical equation shows that 2 moles of Al2O3 produce 4 moles of Al. So, we can set up the following ratio:
2 mol Al2O3 : 4 mol Al
Using the calculated moles of Al2O3 (0.1224 mol), we can determine the moles of Al:
0.1224 mol Al2O3 × (4 mol Al / 2 mol Al2O3) = 0.2448 mol Al
Step 4: Calculate °H for the formation of 0.2448 mol of Al
Given that the enthalpy change for the reaction is 3351.4 kJ for 2 moles of Al2O3, we can set up the following ratio:
2 mol Al2O3 : 3351.4 kJ
Using the calculated moles of Al (0.2448 mol), we can determine the enthalpy change (°H):
0.2448 mol Al × (3351.4 kJ / 2 mol Al2O3) = 410.49 kJ
Therefore, the enthalpy change (°H) for the formation of 12.50 g of aluminum oxide is approximately 410.49 kJ.
To find the enthalpy change (∆H) for the formation of 12.50 g of aluminum oxide (Al2O3), we need to use the given enthalpy change value and the molar mass of aluminum oxide.
First, calculate the number of moles of Al2O3 needed. To do this, divide the given mass (12.50 g) by the molar mass of Al2O3. The molar mass of Al2O3 is calculated by adding the atomic masses of aluminum (Al) and oxygen (O).
The atomic mass of Al = 26.98 g/mol, and the atomic mass of O = 16.00 g/mol. The molar mass of Al2O3 is:
2 × (26.98 g/mol) + 3 × (16.00 g/mol) = 101.96 g/mol.
Now, divide the mass of Al2O3 (12.50 g) by its molar mass (101.96 g/mol) to get the number of moles:
12.50 g / 101.96 g/mol = 0.1227 mol.
Next, use the stoichiometry of the balanced equation to determine the enthalpy change for the given number of moles of Al2O3. The balanced equation is:
2Al2O3 (s) → 4Al (s) + 3O2 (g)
From the balanced equation, we see that 2 moles of Al2O3 produce 4 moles of Al. So, the number of moles of Al produced from 0.1227 mol of Al2O3 is:
0.1227 mol × (4 mol Al / 2 mol Al2O3) = 0.2454 mol Al.
Finally, use the ratio of moles of Al produced to the enthalpy change (∆H) given in the question to find the enthalpy change (∆H) for the formation of 12.50 g of Al2O3:
∆H = 3351.4 kJ × (0.2454 mol / 4 mol) = 204.96 kJ.
Therefore, ∆H for the formation of 12.50 g of aluminum oxide is 204.96 kJ.