Bauxite ore contains aluminum oxide Al2O3(s) which is decomposed using electricity to produce aluminum metal and oxygen gas. What mass of aluminum metal can be procuced from 155g of aluminum oxide?
balance the equation
2Al2O3>>4Al + 3O2
figure the moles of aluminum oxide in 155g
so you get twice those moles of aluminum.
Covert that to grams.
155g of Al203, molar mass is 101.96g/mol, number of moles is 1.52.
ration 2to4 2x=4(1.52)=3.04 moles.
mass = molar mass of Al is 26.98 times number of moles 3.04 = 82.019
To determine the mass of aluminum metal that can be produced from 155g of aluminum oxide, you will need to use its balanced chemical equation and molar masses.
The balanced chemical equation for the decomposition of aluminum oxide is as follows:
2Al2O3(s) → 4Al(s) + 3O2(g)
From the equation, you can see that 2 moles of aluminum oxide are required to produce 4 moles of aluminum. Therefore, the molar ratio between aluminum oxide and aluminum is 2:4, which simplifies to 1:2.
Now, let's calculate the molar mass of aluminum oxide (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 × molar mass of Al) + (3 × molar mass of O)
= (2 × 26.98 g/mol) + (3 × 16.00 g/mol)
= 53.96 g/mol + 48.00 g/mol
= 101.96 g/mol
Next, calculate the number of moles of aluminum oxide in 155g using the formula:
moles = mass / molar mass
moles of Al2O3 = 155g / 101.96 g/mol
≈ 1.52 mol
Since the molar ratio between aluminum oxide and aluminum is 1:2, we can multiply the moles of Al2O3 by 2 to find the moles of aluminum produced.
moles of Al = 1.52 mol × 2
= 3.04 mol
Finally, convert the moles of aluminum to grams using its molar mass:
mass of Al = moles of Al × molar mass of Al
= 3.04 mol × 26.98 g/mol
≈ 82.17g
Therefore, approximately 82.17g of aluminum metal can be produced from 155g of aluminum oxide.
To determine the mass of aluminum metal that can be produced from 155g of aluminum oxide, we need to use stoichiometry.
1. Start by writing a balanced chemical equation for the reaction:
2 Al2O3(s) → 4 Al(s) + 3 O2(g)
2. Identify the molar masses:
Molar mass of Al2O3 = 2(27 g/mol of Al) + 3(16 g/mol of O) = 102 g/mol of Al2O3
Molar mass of Al = 27 g/mol of Al
3. Convert the mass of Al2O3 to moles:
Moles of Al2O3 = Mass of Al2O3 / Molar mass of Al2O3
Moles of Al2O3 = 155g / 102 g/mol of Al2O3
4. Use stoichiometry to find the moles of Al:
Moles of Al = Moles of Al2O3 × (4 mol of Al / 2 mol of Al2O3)
5. Convert moles of Al to mass:
Mass of Al = Moles of Al × Molar mass of Al
Now let's calculate:
Moles of Al2O3 = 155g / 102 g/mol of Al2O3
Moles of Al = (155g / 102 g/mol of Al2O3) × (4 mol of Al / 2 mol of Al2O3)
Mass of Al = (155g / 102 g/mol of Al2O3) × (4 mol of Al / 2 mol of Al2O3) × 27 g/mol of Al
Simplifying the calculation, we find:
Mass of Al = 155g × (4/2) × (27 / 102)
Mass of Al = 155g × (4/2) × (0.265)
Mass of Al = 155g × 2 × 0.265
Mass of Al = 81.23g
Therefore, approximately 81.23 grams of aluminum metal can be produced from 155 grams of aluminum oxide.