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
The molar mass of ferrous sulfide is 87.92 grams and that of iron (III) oxide is 159.7 grams.
To solve this problem, we need to use the balanced equation and the molar masses of the substances involved.
Step 1: Find the molar mass of the compound FeS (ferrous sulfide).
The molar mass of FeS is given as 87.92 grams.
Step 2: Calculate the number of moles of Fe2O3 produced using the given mass of Fe2O3.
To do this, we need to use the molar mass of Fe2O3, which is 159.7 grams.
Number of moles of Fe2O3 = mass of Fe2O3 / molar mass of Fe2O3
Number of moles of Fe2O3 = 0.56 grams / 159.7 grams
Step 3: Use the balanced equation to determine the stoichiometry between FeS and Fe2O3.
According to the balanced equation,
4FeS + 7O2 → 2Fe2O3 + 4SO2
The ratio of FeS to Fe2O3 is 4:2 or simply 2:1.
Step 4: Calculate the number of moles of FeS.
Since the ratio is 2:1, the number of moles of FeS is half the number of moles of Fe2O3.
Number of moles of FeS = Number of moles of Fe2O3 / 2
Step 5: Convert moles of FeS to grams.
To do this, we multiply the number of moles of FeS by its molar mass.
Mass of FeS = Number of moles of FeS * molar mass of FeS
Now you can substitute the values into the equation and calculate it:
Mass of FeS = (0.56 grams / 159.7 grams) / 2 * 87.92 grams
Mass of FeS = 0.0035 grams
Therefore, 0.0035 grams of ferrous sulfide (FeS) will produce 0.56 grams of iron (III) oxide (Fe2O3).
mols Fe2O3 desired = grams/molar mass
Convert mols Fe2O3 to mols FeS using the coefficients in the balanced equation.
g FeS = mols FeS x molar mass FeS