Question

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we, learn, practice, 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 etext.
consider the reaction between reactants s and o₂:
2s(s) + 3o₂(g)→2so₃(g)
correct
use each mole ratio given in the balanced reaction as a conversion factor to determine which reactant is the limiting reactant. the number of moles of so₃ based on s is
7 mol s × \frac{2 mol so₃}{2 mol s} = 7 mol so₃
and the number of moles of so₃ based on o₂ is
9 mol o₂ × \frac{2 mol so₃}{3 mol o₂} = 6.0 mol so₃
since o₂ produces the least amount of product, o₂ is the limiting reactant.
the next step is to determine how much s is consumed by 9 moles of o₂.
consumed moles s = 9 mol o₂ × \frac{2 mol s}{3 mol o₂} = 6 mol s
to find how much excess s remains, subtract the number of moles of s consumed in the reaction from the initial moles of s.
excess mol s = 7 mol s - 6 mol s = 1 mol s
part b
how many moles of o₂ 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 c complete previous part(s)

Explanation:

Step1: Analyze the reaction equation

The balanced reaction is \(2S(s)+3O_{2}(g)\to 2SO_{2}(g)\). The mole ratio of \(S\) to \(O_{2}\) is \(2:3\).

Step2: Determine the limiting reactant

Let's assume we have \(n_{S}\) moles of \(S\) and \(n_{O_{2}}\) moles of \(O_{2}\). From the reaction \(n_{S}/n_{O_{2}}=\frac{2}{3}\). If we have \(7\) moles of \(S\) and \(9\) moles of \(O_{2}\), for \(S\) - based on \(O_{2}\): \(n_{S}^{'}=\frac{2}{3}\times9 = 6\) moles of \(S\) are consumed.

Step3: Calculate the excess \(S\)

Initial moles of \(S = 7\) moles. Consumed moles of \(S=6\) moles. Excess moles of \(S=n_{S}-n_{S}^{'}\)

Answer:

\(1\) mol