Given the equation: 4 Fe + 3 O2 -> 2 Fe2O3
- If I react 5.75 g iron with 3.00 g oxygen, what is the LR?
- What is the theoretical yield of iron(III) oxide?
To do limiting reagent problems you just work two regular stoichiometry problems, which Im sure you know how to do, then comare the products. The SMALLER number wins, like this.
4Fe + 3O2 -> 2Fe2O3
mols Fe = grams/atomic mass =5.75/55.85 = approx 0.1 but you need a more exact number.
Convert this to mols of Fe2O3 that could be produced. mols Fe2O3 produced will be 1/2 mols Fe = approx 0.05.
mols O2 = g/molar mass = 3.00/32 = approx 0.09. Again you need a better number.
Convert to mols Fe2O3 that could be produced. That will be approx 0.09 x 2/3 = approx 0.06 mols Fe2O3.
The smaller number is that of Fe; therefore, Fe is the LR and there will be an excess of O2 Now, ignoring the O2, just calculate how much Fe2O3 will be produced from the 5.75 g Fe. That will be 0.05 mols. Convert to grams by g = mols x molar mass and that is the theoretical yield.
Post your work if you gt stuck.
To find the limiting reactant (LR) in a chemical reaction, you need to compare the ratios of the reactants present with the stoichiometric ratios from the balanced equation.
1. Calculate the moles of iron (Fe) and oxygen (O2):
- Moles of Fe = mass of Fe / molar mass of Fe
- Moles of O2 = mass of O2 / molar mass of O2
The molar masses of Fe and O2 are 55.85 g/mol and 32.00 g/mol, respectively.
2. Find the mole ratio of Fe to O2 from the balanced equation:
- From the equation: 4 Fe + 3 O2 -> 2 Fe2O3
- The stoichiometric ratio of Fe to O2 is 4:3
3. Compare the moles of Fe and O2 using the stoichiometric ratio. Whichever reactant has fewer moles is the limiting reactant.
- Divide the moles of Fe by 4 and the moles of O2 by 3. The reactant with the smaller result is the limiting reactant.
Once you know the limiting reactant, you can determine the theoretical yield of iron(III) oxide (Fe2O3) using the stoichiometry of the balanced equation.
4. Calculate the moles of the limiting reactant.
5. Use the stoichiometry from the balanced equation to find the moles of Fe2O3 that can be formed.
- From the equation: 4 Fe + 3 O2 -> 2 Fe2O3
- The stoichiometric ratio of Fe2O3 to Fe is 2:4
6. Convert the moles of Fe2O3 to grams using the molar mass of Fe2O3, which is 159.69 g/mol.
The theoretical yield of iron(III) oxide is the mass of Fe2O3 calculated in step 6.
Following these steps, you can find the limiting reactant and the theoretical yield of iron(III) oxide.
To determine the limiting reactant (LR), we need to compare the number of moles of each reactant with the stoichiometric coefficient in the balanced equation.
First, we need to calculate the number of moles of each reactant:
Moles of iron (Fe) = mass / molar mass
Moles of oxygen (O2) = mass / molar mass
Using the periodic table, we find that the molar mass of iron (Fe) is approximately 55.85 g/mol and the molar mass of oxygen (O2) is approximately 32.00 g/mol.
For the given values:
Moles of iron (Fe) = 5.75 g / 55.85 g/mol ≈ 0.103 mol
Moles of oxygen (O2) = 3.00 g / 32.00 g/mol ≈ 0.094 mol
Now, we need to compare the moles of each reactant to their stoichiometric coefficient in the balanced equation: 4 Fe + 3 O2 -> 2 Fe2O3
From the balanced equation, we can see that the stoichiometric ratio of Fe to O2 is 4:3. This means that for every 4 moles of Fe, we need 3 moles of O2.
Comparing the moles of reactants:
Moles of Fe / Stoichiometric coefficient of Fe = 0.103 mol / 4 ≈ 0.026 mol Fe per mole of O2
Moles of O2 / Stoichiometric coefficient of O2 = 0.094 mol / 3 ≈ 0.031 mol O2 per mole of O2
Since the moles of Fe per mole of O2 are smaller, Fe is the limiting reactant (LR) in this reaction.
Now, to find the theoretical yield of iron(III) oxide (Fe2O3), we need to use the stoichiometric ratio from the balanced equation.
From the balanced equation, we can see that the stoichiometric ratio of Fe to Fe2O3 is 4:2. This means that for every 4 moles of Fe, we get 2 moles of Fe2O3.
The molar mass of iron(III) oxide (Fe2O3) is approximately 159.69 g/mol.
To calculate the theoretical yield:
Theoretical yield of Fe2O3 = Moles of Fe * Molar mass of Fe2O3 / Stoichiometric coefficient of Fe2O3
Theoretical yield of Fe2O3 = 0.103 mol * 159.69 g/mol / 4 ≈ 4.11 g
Therefore, the theoretical yield of iron(III) oxide (Fe2O3) is approximately 4.11 grams.