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?

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  1. 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.

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

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