A mixture of 10.902 g if iron(II) oxide and
6.524 g of aluminum metal is placed in a
crucible and heated in a high temperature
oven, where a reduction of the oxide takes
place:
3 FeO(s) + 2Al(ℓ) ! 3 Fe(ℓ) + Al2O3(s) .
What is the limiting reactant?
What is the maximum amount of iron that
can be produced?
What is the mass of excess reactant remaining in the crucible?
To determine the limiting reactant, we need to compare the number of moles of each reactant with the stoichiometric coefficients in the balanced chemical equation.
Step 1: Calculate the moles of iron(II) oxide (FeO):
Molar mass of FeO = 55.85 g/mol + 16.00 g/mol = 71.85 g/mol
Moles of FeO = mass / molar mass = 10.902 g / 71.85 g/mol = 0.1517 mol
Step 2: Calculate the moles of aluminum (Al):
Molar mass of Al = 26.98 g/mol
Moles of Al = mass / molar mass = 6.524 g / 26.98 g/mol = 0.2419 mol
Step 3: Use the stoichiometric coefficients from the balanced equation to compare the moles of each reactant:
From the balanced equation, we can see that the ratio of moles between FeO and Al is 3:2.
Since the stoichiometric ratio is 3:2, we can calculate the maximum moles of Fe(ℓ) that can be produced based on the moles of Al:
Maximum moles of Fe(ℓ) = moles of Al × (3 moles of Fe(ℓ) / 2 moles of Al) = 0.2419 mol × (3/2) = 0.3629 mol
Step 4: Compare the calculated maximum moles of Fe(ℓ) with the moles of FeO present:
Since 0.3629 mol of Fe(ℓ) can be produced, and the moles of FeO available is 0.1517 mol, it is clear that FeO is the limiting reactant because there are fewer moles of FeO than required to react completely.
Now let's move on to calculating the maximum amount of iron that can be produced and the mass of excess reactant remaining in the crucible.
Step 5: Calculate the mass of iron that can be produced:
Molar mass of Fe = 55.85 g/mol
Mass of iron produced = moles of Fe(ℓ) × molar mass of Fe = 0.3629 mol × 55.85 g/mol = 20.271 g
Step 6: Calculate the mass of excess reactant remaining:
The excess reactant is aluminum (Al).
Moles of excess Al remaining = moles of Al used - moles of Al required. Since the stoichiometric ratio between FeO and Al is 3:2, the moles of Al required is (3/2) * moles of FeO used.
Moles of Al required = (3/2) * 0.1517 mol = 0.2276 mol
Moles of excess Al remaining = 0.2419 mol - 0.2276 mol = 0.0143 mol
Mass of excess Al remaining = moles of excess Al remaining × molar mass of Al = 0.0143 mol × 26.98 g/mol = 0.386 g
Therefore, the limiting reactant is FeO. The maximum amount of iron that can be produced is 20.271 g, and the mass of excess reactant remaining in the crucible is 0.386 g.
To determine the limiting reactant, we need to compare the number of moles of each reactant with their stoichiometric coefficients in the balanced chemical equation.
First, we calculate the number of moles for each reactant:
- Moles of FeO: Molar mass of FeO = 55.845 g/mol + 15.999 g/mol = 71.844 g/mol
Moles of FeO = mass / molar mass = 10.902 g / 71.844 g/mol ≈ 0.152 mol
- Moles of Al: Molar mass of Al = 26.982 g/mol
Moles of Al = mass / molar mass = 6.524 g / 26.982 g/mol ≈ 0.242 mol
Next, we compare the moles of each reactant to their stoichiometric coefficients in the balanced chemical equation: FeO:Al = 3:2
- Moles of FeO available: 0.152 mol
- Moles of Al available: 0.242 mol
Since the stoichiometric ratio between FeO and Al is 3:2, we can see that we have an excess of Al. Therefore, Al is the excess reactant and FeO is the limiting reactant.
To find the maximum amount of iron that can be produced, we need to calculate the moles of Fe that can be produced based on the balanced chemical equation: 3 FeO:3 Fe
Since 3 moles of FeO produce 3 moles of Fe, the number of moles of Fe produced will be the same as the moles of FeO used. Therefore, the maximum amount of iron that can be produced is 0.152 mol.
To find the mass of the excess reactant remaining in the crucible, we first calculate the moles of excess Al reacted by subtracting the moles of Al that reacted from the initial moles of Al available:
Moles of excess Al = Moles of Al available - Moles of Al reacted
= 0.242 mol - 0.152 mol (since FeO is the limiting reactant)
= 0.09 mol
Finally, we calculate the mass of excess Al remaining by multiplying the moles of excess Al by its molar mass:
Mass of excess Al = Moles of excess Al x Molar mass of Al
≈ 0.09 mol x 26.982 g/mol
≈ 2.429 g
Therefore, the mass of the excess reactant remaining in the crucible is approximately 2.429 grams.