If 19:3 of iron ore (Fe2O3) reacts completely with carbon monoxide . Calculate the mass of iron Fe that is formed
To calculate the mass of iron (Fe) formed when 19:3 of iron ore (Fe2O3) reacts completely with carbon monoxide (CO), we need to follow a few steps:
1. Determine the molar mass of Fe2O3 and Fe.
- The molar mass of Fe2O3 is calculated as follows:
- Molar mass of Fe = atomic mass of Fe = 55.845 g/mol
- Molar mass of O = atomic mass of O = 16.00 g/mol
- Therefore, molar mass of Fe2O3 = (2 * 55.845) + (3 * 16.00) = 159.69 g/mol
- The molar mass of Fe is simply the atomic mass of Fe, which is 55.845 g/mol.
2. Calculate the moles of Fe2O3:
- Given the ratio of 19:3, we need to find the moles of Fe2O3 that correspond to the given amount.
- Assume we have 19 moles of Fe2O3. Then we can calculate the moles of Fe2O3 as follows:
- Moles of Fe2O3 = 19 moles
3. Determine the stoichiometry of the reaction between Fe2O3 and CO:
- The balanced equation for the reaction of Fe2O3 with CO is:
- 3Fe2O3 + CO → 2Fe3O4 + CO2
- From the balanced equation, we can see that 3 moles of Fe2O3 react to produce 2 moles of Fe.
- So, the molar ratio of Fe2O3 to Fe is 3:2.
4. Calculate the moles of Fe using the molar ratio:
- Moles of Fe = (moles of Fe2O3 * 2) / 3
= (19 * 2) / 3
= 38 / 3
= 12.67 moles
5. Calculate the mass of Fe formed:
- Mass of Fe = Moles of Fe * Molar mass of Fe
= 12.67 moles * 55.845 g/mol
= 707.81 grams (rounded to two decimal places)
Therefore, the mass of iron (Fe) formed when 19:3 of iron ore (Fe2O3) reacts completely with carbon monoxide (CO) is approximately 707.81 grams.