6Na+Fe2O3=3Na2O+2Fe
if Interact 100g of Na with 100g of Fe2O3,Calculate:
- The resulting mass of iron?
This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants.
6Na + Fe2O3 ==> 3Na2O + 2Fe
1. mols Na = grams/molar mass
2. mols Fe2O3 = grams/molar mass
3a. Using the coefficients in the balanced equation, convert mols Na to mols of the product, Fe.
3b. Do the same for mols Fe2O3 to mols Fe.
3c. It is likely that the values for mols Fe from 3a and 3b will not agree which means one of them is not right. The correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
4. Now convert mols Fe to g. g = mols x molar mass.
To calculate the resulting mass of iron, we need to determine which reactant is the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.
1. Calculate the molar masses of the reactants:
- Sodium (Na) = 22.99 g/mol
- Iron(III) oxide (Fe2O3) = 159.69 g/mol
2. Convert the given masses of Na and Fe2O3 into moles:
- Moles of Na = Mass of Na / Molar mass of Na = 100 g / 22.99 g/mol ≈ 4.35 mol
- Moles of Fe2O3 = Mass of Fe2O3 / Molar mass of Fe2O3 = 100 g / 159.69 g/mol ≈ 0.626 mol
3. Write and balance the balanced equation to determine the stoichiometry of the reaction:
6Na + Fe2O3 -> 3Na2O + 2Fe
According to the equation, 6 moles of Na react with 1 mole of Fe2O3 to produce 2 moles of Fe.
4. Determine the limiting reactant. To do this, compare the mole ratios from the balanced equation to the actual moles of each reactant. The reactant with the smaller ratio will be the limiting reactant.
- Na:Fe2O3 ratio is 6:1
- Actual ratio is 4.35:0.626
Since 0.626 < 4.35, Fe2O3 is the limiting reactant.
5. Calculate the number of moles of Fe that can be obtained from Fe2O3:
From the balanced equation, 1 mole of Fe2O3 produces 2 moles of Fe.
Moles of Fe = 0.626 mol of Fe2O3 × (2 mol of Fe / 1 mol of Fe2O3) = 1.252 mol of Fe
6. Calculate the resulting mass of iron:
Resulting mass of Fe = Moles of Fe × Molar mass of Fe = 1.252 mol × 55.85 g/mol ≈ 69.99 g
Therefore, the resulting mass of iron is approximately 69.99 grams.
To calculate the resulting mass of iron, we first need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the amount of product that can be formed.
Let's find the moles of Na and Fe2O3:
Molar mass of Na (sodium) = 22.99 g/mol
Molar mass of Fe2O3 (iron(III) oxide) = 159.69 g/mol
Moles of Na = mass / molar mass = 100 g / 22.99 g/mol = 4.35 mol
Moles of Fe2O3 = mass / molar mass = 100 g / 159.69 g/mol = 0.626 mol
From the balanced equation, we can see that the stoichiometric ratio between Na and Fe2O3 is 6:1. This means that 6 moles of Na react with 1 mole of Fe2O3 to produce 2 moles of Fe.
Since we have 4.35 moles of Na and 0.626 moles of Fe2O3, we need to check which one is the limiting reactant. Dividing the moles of each reactant by their stoichiometric coefficient, we can compare the ratios:
Moles of Na / stoichiometric coefficient of Na = 4.35 mol / 6 = 0.725 mol
Moles of Fe2O3 / stoichiometric coefficient of Fe2O3 = 0.626 mol / 1 = 0.626 mol
We can see that the moles of Na (0.725 mol) are greater than the moles of Fe2O3 (0.626 mol). Therefore, Fe2O3 is the limiting reactant.
From the balanced equation, we can see that for every 1 mole of Fe2O3, we get 2 moles of Fe. Therefore, we can convert the moles of Fe2O3 to moles of Fe:
Moles of Fe = moles of Fe2O3 * (2 moles of Fe / 1 mole of Fe2O3) = 0.626 mol * (2/1) = 1.252 mol
Finally, we can calculate the mass of iron using the moles of Fe and the molar mass of iron:
Molar mass of Fe (iron) = 55.85 g/mol
Mass of Fe = moles of Fe * molar mass of Fe = 1.252 mol * 55.85 g/mol = 69.95 g
Therefore, the resulting mass of iron is 69.95 grams.