In the following reaction, how many grams of ferrous sulfide (FeS) will produce 0.56 grams of iron (III) oxide (Fe2O3)?
4FeS + 7O2 d 2Fe2O3 + 4SO2
.56g of Fe2O3 = .56/159.69 = 0.0035 moles
.0035 moles FeS = 0.307g FeS
To find the amount of ferrous sulfide (FeS) needed to produce 0.56 grams of iron (III) oxide (Fe2O3), we need to use stoichiometry.
1. Start by writing down the balanced chemical equation:
4FeS + 7O2 → 2Fe2O3 + 4SO2
2. Calculate the molar mass of iron (III) oxide (Fe2O3) and ferrous sulfide (FeS).
The molar mass of Fe2O3 = (2 * 55.85 g/mol) + (3 * 16.00 g/mol) = 159.69 g/mol
The molar mass of FeS = 55.85 g/mol + 32.06 g/mol = 87.91 g/mol
3. Determine the moles of iron (III) oxide (Fe2O3) using its molar mass:
Moles of Fe2O3 = Mass of Fe2O3 / Molar mass of Fe2O3
Moles of Fe2O3 = 0.56 g / 159.69 g/mol
4. Use the stoichiometric ratio between Fe2O3 and FeS to find the moles of FeS required:
According to the balanced equation, 4 moles of FeS react with 2 moles of Fe2O3.
This means that the ratio of moles of FeS to Fe2O3 is 4:2 or 2:1.
Moles of FeS = (Moles of Fe2O3) * (2 moles of FeS / 2 moles of Fe2O3)
Moles of FeS = 0.56 g / 159.69 g/mol * (2 mol FeS / 2 mol Fe2O3)
5. Calculate the mass of FeS using its molar mass and the moles of FeS:
Mass of FeS = Moles of FeS * Molar mass of FeS
Mass of FeS = Moles of FeS * 87.91 g/mol
By following these steps, you should be able to find the amount of ferrous sulfide (FeS) needed to produce 0.56 grams of iron (III) oxide (Fe2O3).