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

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Answer 1 · 2018-01-20T10:25:03

$=80.09g$

Write the balanced equation as presented.

$4FeS_2+11O_2->2Fe_2O_3+8SO_2$
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.

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

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

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

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