Fe2O3 + 3CO= 2Fe + 3CO2, SUPPOSE THAT 1.64kg OF Fe ARE OBTAINED FROM A 2.62kg SAMPLE OF Fe2O3(ORE). ASSUMING THE REACTION IS COMPLETE, WHAT IS THE PERCENT PURITY OF THE Fe2O3 IN THE ORIGINAL SAMPLE?
one needs one mole of ironIIoxide to make two moles of iron.
MolesIron obtained: 1640/atomicmassFe
molesFe2O3: one half that.
grams Fe2O3: moles*molmass
percent purity: gramsFe2O3/2620
85.12 grams i think .
To find the percent purity of Fe2O3 in the original sample, we need to compare the mass of Fe obtained with the expected theoretical yield.
Step 1: Calculate the molar masses of Fe2O3 and Fe.
- The molar mass of Fe2O3:
Fe = 55.845 g/mol
O = 16.00 g/mol (x3)
Fe2O3 = 55.845 g/mol + (16.00 g/mol x 3) = 159.69 g/mol
- The molar mass of Fe: 55.845 g/mol
Step 2: Calculate the number of moles of Fe obtained.
- The balanced equation shows that 2 moles of Fe are obtained from 1 mole of Fe2O3. Therefore, the number of moles of Fe obtained can be calculated as follows:
Moles of Fe = Mass of Fe / Molar mass of Fe
Moles of Fe = 1.64 kg x (1000 g/kg) / 55.845 g/mol = 29.38 mol
Step 3: Calculate the number of moles of Fe2O3 in the original sample.
- The balanced equation shows that the molar ratio of Fe2O3 to Fe is 1:2. Therefore, the number of moles of Fe2O3 in the original sample can be calculated as follows:
Moles of Fe2O3 = (Moles of Fe / 2) = 29.38 mol / 2 = 14.69 mol
Step 4: Calculate the mass of Fe2O3 in the original sample.
- The mass of Fe2O3 in the original sample can be calculated using the molar mass formula:
Mass of Fe2O3 = Moles of Fe2O3 x Molar mass of Fe2O3
Mass of Fe2O3 = 14.69 mol x 159.69 g/mol = 2342.86 g
Step 5: Calculate the percent purity of Fe2O3.
- The percent purity of Fe2O3 in the original sample can be calculated as follows:
Percent purity = (Mass of pure Fe2O3 / Mass of original sample) x 100%
Percent purity = (2342.86 g / 2620 g) x 100% ≈ 89.41%
Therefore, the percent purity of Fe2O3 in the original sample is approximately 89.41%.
To find the percent purity of Fe2O3 in the original sample, you need to compare the mass of the actual Fe obtained to the theoretical maximum yield of Fe that could be obtained if all the Fe2O3 reacted.
First, let's calculate the molar mass of Fe2O3 and Fe. The molar masses are:
- Molar mass of Fe2O3 = 2 * atomic mass of Fe + 3 * atomic mass of O
- Molar mass of Fe = atomic mass of Fe
Using the atomic masses from the periodic table, let's calculate them:
- Atomic mass of Fe = 55.845 g/mol
- Atomic mass of O = 16.00 g/mol
Therefore,
- Molar mass of Fe2O3 = 2 * 55.845 g/mol + 3 * 16.00 g/mol = 159.69 g/mol
- Molar mass of Fe = 55.845 g/mol
Now we can convert the masses given to moles:
- Moles of Fe = 1.64 kg * (1000 g/kg) / 55.845 g/mol = 29.3 mol
- Moles of Fe2O3 = 2.62 kg * (1000 g/kg) / 159.69 g/mol = 16.4 mol
According to the balanced equation, the stoichiometry between Fe2O3 and Fe is 1:2, which means 1 mole of Fe2O3 yields 2 moles of Fe.
Therefore, if all the Fe2O3 reacted completely, the maximum theoretical yield of Fe would be:
- Moles of Fe2O3 * (2 moles of Fe / 1 mole of Fe2O3) = 16.4 mol * (2/1) = 32.8 mol
Now we can calculate the percent purity using the actual yield and the theoretical yield:
- Percent purity = (Actual yield of Fe / Theoretical yield of Fe) * 100
- Percent purity = (29.3 mol / 32.8 mol) * 100
Lastly, calculate the percentage:
- Percent purity = 89.33%
Therefore, the percent purity of Fe2O3 in the original sample is approximately 89.33%.