One of the reactions that occurs in a blast furnace, in which iron ore is converted to cast iron, is
Fe2O3 + 3CO → 2Fe + 3CO2
Suppose that 1.59 × 103 kg of Fe is obtained from a 2.54 × 103 kg sample of Fe2O3. Assuming that the reaction goes to completion, what is the percent purity of Fe2O3 in the original sample?
There are easier ways to do this but I think this is the best way to go about it. I would change the kg to grams.
mols Fe = grams Fe/atomic mass Fe = ?
Using the coefficients in the balanced equation, convert mols Fe to mols Fe2O3. That is mols Fe x (1 mol Fe2O3/2 mols Fe) = ?
Now convert mols Fe2O3 to grams Fe2O3.
g Fe2O3 = mols Fe2O3 x molar mass Fe2O3 = ?
Finally, %Fe2O3 = %purity = (g Fe2O3/
%Fe = (mass Fe/mass sample)*100 = ?
Substitute and solve for %Fe which is the percent purity.
Can you put the figures into the equation for me, because I do not quite understand this?
To find the percent purity of Fe2O3 in the original sample, you need to compare the amount of Fe obtained from the reaction to the theoretical maximum amount of Fe that could be obtained if all the Fe2O3 reacted.
First, calculate the molar mass of Fe2O3:
Molar mass of Fe = 55.85 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of Fe2O3 = (2 × Molar mass of Fe) + (3 × Molar mass of O)
= (2 × 55.85 g/mol) + (3 × 16.00 g/mol)
= 159.69 g/mol
Now, calculate the number of moles of Fe obtained from the reaction:
Mass of Fe obtained = 1.59 × 10^3 kg = 1.59 × 10^6 g
Moles of Fe obtained = Mass of Fe obtained / Molar mass of Fe
= 1.59 × 10^6 g / 55.845 g/mol
≈ 28,500 mol
According to the balanced equation, the stoichiometry of Fe to Fe2O3 is 2:1. This means that for every 2 moles of Fe, 1 mole of Fe2O3 is required.
Therefore, moles of Fe2O3 required = (1/2) × Moles of Fe obtained
= (1/2) × 28,500 mol
= 14,250 mol
Finally, calculate the mass of Fe2O3 in the original sample:
Mass of Fe2O3 in the original sample = Moles of Fe2O3 required × Molar mass of Fe2O3
= 14,250 mol × 159.69 g/mol
≈ 2.277 × 10^6 g
Now, to find the percent purity of Fe2O3, divide the mass of the Fe2O3 in the original sample by the mass of the sample and multiply by 100:
Percent purity of Fe2O3 = (Mass of Fe2O3 in the original sample / Mass of the sample) × 100
= (2.277 × 10^6 g / 2.54 × 10^6 g) × 100
≈ 89.7%
Therefore, the percent purity of Fe2O3 in the original sample is approximately 89.7%.