SO3 can be produced in the following two-step process:

FeS2 + O2 ---> Fe2O3 + SO2
S02 + O2 ----> SO3

Assuming that all the FeS2 reacts, how many grams of SO3 are produced when 20.0 g of the FeS2 reacts with 16.0 g of O2?

The equations are not balanced.

1. Balance the equations.
2. Since BOTH reactants are given, you must recognize that this is a limiting reagent problem.
3. Convert 20 g FeS2 to moles. Moles = g/molar mass.
4. Convert 16 g O2 to moles. Same formula.
5. Using the coefficients in the balanced equation, convert moles of FeS2 in step 3 to moles SO2.
6. Do the same thing for moles O2 in step 4 and convert to moles SO2.
7. Obviously, both 5 and 6 can't be right (assuming the answer for moles SO2 is different) so you must pick the one that is correct. Pick the SMALLER of the moles in step 5 and 6. That is the limiting reagent.
8. Use that number of moles SO2 as the starting point in the second equation, and go through the steps to convert moles SO2 to moles SO3. (This part is not a limiting reagent problem.)
9. Now convert moles SO3 to grams SO3. Post your work if you get stuck.

I'm quite confused with this process, by doing this I get approx. 26 but the actual answer should be approx. 6 :(

To determine the amount of SO3 produced, we need to calculate the limiting reactant and use its stoichiometric ratio with SO3.

First, let's calculate the moles of FeS2 and O2 using their respective molar masses:

Molar mass of FeS2 = 55.85 g/mol (Fe) + 32.06 g/mol (S) x 2 = 119.98 g/mol

Moles of FeS2 = mass of FeS2 / molar mass of FeS2 = 20.0 g / 119.98 g/mol

Molar mass of O2 = 16.00 g/mol x 2 = 32.00 g/mol

Moles of O2 = mass of O2 / molar mass of O2 = 16.0 g / 32.00 g/mol

Now, let's determine the theoretical yield of Fe2O3 using the stoichiometric ratio between FeS2 and Fe2O3:

From the balanced equation: FeS2 + O2 ---> Fe2O3 + SO2

The ratio of FeS2 : Fe2O3 is 1:1. This means that for every 1 mole of FeS2, we get 1 mole of Fe2O3.

Since all the FeS2 reacts, the moles of Fe2O3 produced is equal to the moles of FeS2: Moles of Fe2O3 = Moles of FeS2

Next, let's determine the theoretical yield of SO3 using the stoichiometric ratio between O2 and SO3:

From the balanced equation: SO2 + O2 ----> SO3

The ratio of O2 : SO3 is 1:1. This means that for every 1 mole of O2, we get 1 mole of SO3.

Since all the O2 reacts, the moles of SO3 produced is equal to the moles of O2: Moles of SO3 = Moles of O2

Finally, let's calculate the mass of SO3 produced using its molar mass:

Molar mass of SO3 = 32.06 g/mol (S) + 16.00 g/mol (O) x 3 = 80.06 g/mol

Mass of SO3 = Moles of SO3 * Molar mass of SO3 = Moles of O2 * 80.06 g/mol

Now, let's substitute the values:

Moles of FeS2 = 20.0 g / 119.98 g/mol = 0.1667 mol
Moles of O2 = 16.0 g / 32.00 g/mol = 0.5000 mol

Moles of Fe2O3 = Moles of FeS2 = 0.1667 mol
Moles of SO3 = Moles of O2 = 0.5000 mol

Mass of SO3 = Moles of SO3 * Molar mass of SO3 = 0.5000 mol * 80.06 g/mol = 40.03 g

Therefore, when 20.0 g of FeS2 reacts with 16.0 g of O2, the mass of SO3 produced is 40.03 grams.

To determine the grams of SO3 produced, we need to calculate the number of moles of FeS2 and O2 used in the reaction and then use the stoichiometry to find the moles of SO3 produced. Finally, we can convert moles to grams.

Step 1: Calculate the moles of FeS2 and O2:
Molar mass of FeS2 = 119.98 g/mol
Molar mass of O2 = 32.00 g/mol

moles of FeS2 = mass of FeS2 / molar mass of FeS2
= 20.0 g / 119.98 g/mol
= 0.167 moles

moles of O2 = mass of O2 / molar mass of O2
= 16.0 g / 32.00 g/mol
= 0.50 moles

Step 2: Use stoichiometry to find moles of SO3 produced:
From the balanced equation, we see that the ratio of FeS2 to SO3 is 1:1, and the ratio of O2 to SO3 is 1:1.

Since all 0.167 moles of FeS2 react, the same amount of moles of SO3 is produced.

Step 3: Convert moles to grams:
Molar mass of SO3 = 80.06 g/mol

grams of SO3 = moles of SO3 × molar mass of SO3
= 0.167 moles × 80.06 g/mol
= 13.35 g

Therefore, when 20.0 g of FeS2 reacts with 16.0 g of O2, 13.35 grams of SO3 are produced.