Question

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ae limiting reactant and theoretical yield from initial moles of reactants. read
section 8.5. you can click on the review link to access the section in your e text.
consider the reaction between reactants s and o₂:
2s(s) + 3o₂(g)→2so₃(g)
part a
if a reaction vessel initially contains 7 mol s and 9 mol o₂, how many moles of s will be in the reaction vessel once the reactants have reacted as much as possible? (assume 100% actual yield)
express your answer using two significant figures.

part b
complete previous part(s)
part c
complete previous part(s)
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Explanation:

Step1: Determine the limiting reactant

From the balanced equation \(2S(s)+3O_{2}(g)\to 2SO_{3}(g)\), the mole ratio of \(S\) to \(O_{2}\) is \(2:3\).
If we have \(n(S) = 7\space mol\) and \(n(O_{2})=9\space mol\).
For \(S\): if all \(S\) reacts, the moles of \(O_{2}\) required \(n_{req}(O_{2})=\frac{3}{2}\times n(S)=\frac{3}{2}\times7 = 10.5\space mol\).
For \(O_{2}\): if all \(O_{2}\) reacts, the moles of \(S\) required \(n_{req}(S)=\frac{2}{3}\times n(O_{2})=\frac{2}{3}\times9=6\space mol\).
Since \(6\space mol\) of \(S\) is required to react with \(9\space mol\) of \(O_{2}\), \(O_{2}\) is the limiting reactant.

Step2: Calculate the moles of \(S\) that react

Using the mole - ratio from the balanced equation.
The mole ratio of \(S\) to \(O_{2}\) is \(2:3\). Let \(x\) be the moles of \(S\) that react.
\(\frac{x}{9}=\frac{2}{3}\), so \(x = 6\space mol\)

Step3: Calculate the moles of \(S\) remaining

The initial moles of \(S\) is \(n_{initial}(S)=7\space mol\).
The moles of \(S\) remaining \(n_{remaining}(S)=n_{initial}(S)-x\)
\(n_{remaining}(S)=7 - 6=1\space mol\)

Answer:

\(1\space mol\)