Question

4FeS2 + 11 O2­ → 2 Fe2O3 + 8 SO2 Given 75.0 grams of FeS2, how many grams of sulfur dioxide are produced (MM in g/mol: FeS2=119.99, O2=32, Fe2O3=159.7, SO2=64.07) ? 40.0. g 70.0 g 150. g 80.0 g

Answer 1

`=80.09g`

Explanation:

1. Write the balanced equation as presented.

   `4FeS_2+11O_2->2Fe_2O_3+8SO_2`

2. Given the mass of `FeS_2`, per convention will be converted to mol. To do that, find the molar mass of the compound which is obtainable from the periodic table where the equivalence statement `1molFeS_2-=119.98gFeS_2` is derived.

   `etaFeS_2=75.0cancel(g)xx(1mol)/(119.99cancel(g))`
   `etaFeS_2=0.625mol`

3. Now, referring to the balanced equation, take the mole ratio of the `FeS_2` against the `SO_2` which is found to be `4:8`. This value provides the relationship that converts `etaFeS_2` to `etaSO_2`; thus,

   `etaSO_2=0.625cancel(molFeS_2)xx(8molSO_2)/(4cancel(molFeS_2))`
   `etaSO_2=1.250mol`

4. Finally, the problem is asking for the mass of `SO_2` produced in the reaction. This case, molar mass is needed to again convert the value from mole to mass as shown below.

   `mSO_2=1.250cancel(molSO_2)xx(64.07gSO_2)/(1cancel(molSO_2))`
   `mSO_2=80.09g`