Question

4 fes(s) + 7 o₂(g) → 2 fe₂o₃(s) + 4 so₂(g)
due to the presence of fes(s) as an impurity, the combustion of some types of coal results
in the formation of so₂(g), as represented by the equation above. also, so₂(g) can react
with o₂(g) to form so₃(g) as represented by the equation below
so₂(g) + o₂(g) ⇌ so₃(g) + o₂(g)
δh°_rxn = -242 kj/mol_rxn, δs°_rxn = -25 j/(k·mol_rxn)
which of the following statements is correct when 2.0 mol of fes(s) reacts with 4.0 mol
of o₂(g)?
a o₂(g) is the limiting reactant and 2.0 mol of so₂(g) is formed
b o₂(g) is the limiting reactant and 4.0 mol of so₂(g) is formed
c fes(s) is the limiting reactant and 4.0 mol of so₂(g) is
formed
d fes(s) is the limiting reactant and 0.5 mol of o₂(g) remains
unreacted

Explanation:

Step1: Determine the limiting reactant

From the balanced equation \(4FeS(s)+7O_{2}(g)\to2Fe_{2}O_{3}(s)+4SO_{2}(g)\), the mole ratio of \(FeS\) to \(O_{2}\) is \(4:7\).
Given \(n(FeS) = 2.0\ mol\) and \(n(O_{2})=4.0\ mol\).
If \(FeS\) is completely consumed, the moles of \(O_{2}\) required \(n_{req}(O_{2})=\frac{7}{4}\times n(FeS)\).
Substitute \(n(FeS) = 2.0\ mol\), we get \(n_{req}(O_{2})=\frac{7}{4}\times2.0\ mol = 3.5\ mol\).
Since \(n(O_{2}) = 4.0\ mol>3.5\ mol\), \(FeS\) is the limiting reactant.

Step2: Calculate the moles of \(SO_{2}\) formed

From the mole ratio of \(FeS\) to \(SO_{2}\) (\(4:4 = 1:1\)) in the balanced equation \(4FeS(s)+7O_{2}(g)\to2Fe_{2}O_{3}(s)+4SO_{2}(g)\).
If \(n(FeS) = 2.0\ mol\), then \(n(SO_{2})=2.0\ mol\) (based on the stoichiometry).
The moles of \(O_{2}\) unreacted \(n_{unreacted}(O_{2})=n(O_{2})-n_{req}(O_{2})\)
\(n_{unreacted}(O_{2})=4.0\ mol - 3.5\ mol=0.5\ mol\)

Answer:

D. \(FeS(s)\) is the limiting reactant and \(0.5\ mol\) of \(O_{2}(g)\) remains unreacted.