f 147 grams of FeS2 is allowed to react with 88 grams of O2 according to the following unbalanced equation, how many grams of Fe2O3 are produced?
FeS2 + O2 → Fe2O3 + SO2
This is a limiting reagent. You know that because BOTH of the reactant are given.
First, balance the equation.
Convert 147 g FeS2 to moles. moles = grams/molar mass.
Convert 88 g oxygen to moles.
Using the coefficients in the balanced equation, convert moles FeS2 to moles Fe2O3.
Same procedure, convert moles oxygen to moles Fe2O3.
It is more than likely that the two numbers for moles Fe2O3 will not agree which means one of them is wrong. In limiting reagent problems, the correct answer is ALWAYS the smaller value and the reactant providing that number is the limiting reagent.
Using the smaller value, convert to grams Fe2O3. g = moles x molar mass
Post your work if you get stuck.
319
1277
I got 36.57g Fe2O3, what is the correct answer?
80.0
Well, let's balance the equation and see what happens when we mix some chemistry with a little bit of comedy!
2FeS2 + 11O2 → Fe2O3 + 8SO2
Now that we have a balanced equation, we can use stoichiometry to find the grams of Fe2O3 produced.
First, let's determine the limiting reactant by comparing the molar ratios.
The molar mass of FeS2 is 87.92 g/mol, while the molar mass of O2 is about 32 g/mol.
So, for FeS2: 147 g / 87.92 g/mol = 1.67 mol
And for O2: 88 g / 32 g/mol ≈ 2.75 mol
Looking at the balanced equation, we can see that it takes 2 moles of FeS2 to produce 1 mole of Fe2O3. Since we have 1.67 moles of FeS2, we can expect to produce 1.67 / 2 = 0.84 moles of Fe2O3.
Finally, we can calculate the grams of Fe2O3 by multiplying the moles by the molar mass of Fe2O3, which is 159.69 g/mol.
Grams of Fe2O3 = 0.84 mol × 159.69 g/mol = 134.11 g
So, approximately 134.11 grams of Fe2O3 are produced.
Hope that answers your question with a touch of laughter!
To solve this problem, we need to follow a few steps:
Step 1: Write down the given information and balance the equation.
Given:
Mass of FeS2 = 147 grams
Mass of O2 = 88 grams
Unbalanced Equation:
FeS2 + O2 → Fe2O3 + SO2
Balanced Equation:
4FeS2 + 11O2 → 2Fe2O3 + 8SO2
Step 2: Determine the molar masses of the compounds involved.
Molar Mass:
FeS2 = (1 atom of Fe × atomic mass of Fe) + (2 atoms of S × atomic mass of S)
= (1 × 55.845) + (2 × 32.06)
= 119.77 g/mol
O2 = 2(atomic mass of O)
= 2(16.00)
= 32.00 g/mol
Fe2O3 = (2 atoms of Fe × atomic mass of Fe) + (3 atoms of O × atomic mass of O)
= (2 × 55.845) + (3 × 16.00)
= 159.69 g/mol
Step 3: Convert the given masses to moles.
Moles of FeS2 = Mass of FeS2 / Molar Mass of FeS2
= 147 g / 119.77 g/mol
≈ 1.23 mol
Moles of O2 = Mass of O2 / Molar Mass of O2
= 88 g / 32.00 g/mol
≈ 2.75 mol
Step 4: Use the balanced equation to determine the stoichiometry between reactants and products.
From the balanced equation, we can see that the ratio of FeS2 to Fe2O3 is 4:2. Therefore, for every 4 moles of FeS2, we produce 2 moles of Fe2O3.
Step 5: Calculate the moles of Fe2O3 produced.
Moles of Fe2O3 = (Moles of FeS2 / stoichiometric coefficient of FeS2) × stoichiometric coefficient of Fe2O3
= (1.23 mol / 4) × 2
≈ 0.615 mol
Step 6: Convert moles of Fe2O3 to grams.
Mass of Fe2O3 = Moles of Fe2O3 × Molar Mass of Fe2O3
= 0.615 mol × 159.69 g/mol
≈ 98.02 g
Therefore, approximately 98.02 grams of Fe2O3 are produced when 147 grams of FeS2 reacts with 88 grams of O2.