Fe2O3(s) + 2Al(s)-> 2Fe(l) + Al2O3(s)
What is the theoretical yield of iron if we begin the reaction with 10.5 g of Al (assume an excess of Fe2O3)
mols Al = grams/atomic mass = ?
Using the coefficients in the balanced equation, convert mols Al to mols Fe. That is the theoretical yield. If you want it in grams then g = mols x atomic mass.
5.6 × 102 kJ
To find the theoretical yield of iron in this reaction, you need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
The balanced equation shows that the stoichiometric ratio between Fe2O3 and Al is 1:2. This means that for every 1 mole of Fe2O3, 2 moles of Al are required.
1. Convert the mass of Al to moles:
molar mass of Al = 26.98 g/mol
moles of Al = mass of Al / molar mass of Al
moles of Al = 10.5 g / 26.98 g/mol
moles of Al ≈ 0.389 mol
2. Use the stoichiometric ratio between Fe2O3 and Al to determine the moles of Fe2O3 needed:
moles of Fe2O3 = 0.389 mol of Al × (1 mol of Fe2O3 / 2 mol of Al)
moles of Fe2O3 ≈ 0.1945 mol
Since there is an excess of Fe2O3, the moles of Fe2O3 will not be completely consumed. Therefore, the limiting reactant is Fe2O3.
3. Use the stoichiometric ratio between Fe2O3 and Fe to determine the theoretical yield of iron:
moles of Fe = 0.1945 mol of Fe2O3 × (2 mol of Fe / 1 mol of Fe2O3)
moles of Fe ≈ 0.389 mol
4. Convert the moles of Fe to grams:
molar mass of Fe = 55.85 g/mol
mass of Fe = moles of Fe × molar mass of Fe
mass of Fe = 0.389 mol × 55.85 g/mol
mass of Fe ≈ 21.67 g
Therefore, the theoretical yield of iron in this reaction is approximately 21.67 grams.
To calculate the theoretical yield of iron in this reaction, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
In this case, we have 10.5 g of Al and an excess of Fe2O3. The molar mass of Al is 26.98 g/mol. To find the moles of Al, we can use the formula:
moles of Al = mass of Al / molar mass of Al
moles of Al = 10.5 g / 26.98 g/mol
moles of Al = 0.389 mol
According to the balanced chemical equation, the stoichiometric ratio between Al and Fe is 2:2, which means that for every 2 moles of Al, we should obtain 2 moles of Fe.
By looking at the equation, we see that the stoichiometric ratio between Fe2O3 and Fe is 1:2. This means that for every 1 mole of Fe2O3, we should obtain 2 moles of Fe.
Since the ratio between Al and Fe is the same as that between Fe2O3 and Fe, the moles of Fe formed will be equal to the moles of Al used.
Therefore, the theoretical yield of Fe is 0.389 mol.
To calculate the mass of Fe, we need to use the formula:
mass of Fe = moles of Fe * molar mass of Fe
The molar mass of Fe is 55.85 g/mol.
mass of Fe = 0.389 mol * 55.85 g/mol
mass of Fe = 21.70 g
Hence, the theoretical yield of iron is 21.70 grams if we begin the reaction with 10.5 g of Al.