Answer:
[tex]\large \boxed{56.1 \, \% }[/tex]
Explanation:
We have the masses of two reactants, so this is a limiting reactant problem.
We will need a balanced equation with masses and molar masses of the compounds involved. Gather all the information in one place
Mᵣ: 64.06 32.00 80.06
2SO₂ + O₂ ⟶ 2SO₃
m/g: 544.5 160.0
1. Theoretical yield
(a) Calculate the moles of each reactant
[tex]\text{Moles of SO}_{2} = \text{544.5 g} \times \dfrac{\text{1 mol}}{\text{64.06 g }} = \text{8.500 mol}\\\\\text{Moles of O} _{2}= \text{160.0 g} \times \dfrac{\text{1 mol}}{\text{32.00 g}} = \text{5.000 mol}[/tex]
(b) Identify the limiting reactant
Calculate the moles of SO₃ we can obtain from each reactant.
From SO₂:
The molar ratio of SO₂ to SO₃ is 2:2
[tex]\text{Moles of SO}_{3} = \text{8.500 mol SO}_{2} \times \dfrac{\text{2 mol SO} _{3}}{\text{2 mol SO}_{2}} = \text{8.500 mol SO}_{3}[/tex]
From O₂ :
The molar ratio of SO₂:O₂ is 2 mol O₂:1 mol O₂.
[tex]\text{Moles of SO$_{3}$} = \text{5.000 mol O}_{2} \times \dfrac{\text{2 mol SO$_{3}$}}{\text{1 mol O$_{2}$}} = \text{10.00 mol SO$_{3}$}[/tex]
The limiting reactant is SO₂ because it gives the smaller amount of SO₃.
(c) Calculate the theoretical yield of SO₃.
[tex]\text{Theor. yield of SO}_{3} = \text{8.500 mol} \times \dfrac{\text{80.06 g}}{\text{ 1 mol}} = \text{680.5 g}[/tex]
2. Calculate the percent yield of SO₃
[tex]\text{\% yield} = \dfrac{\text{ 382.0 g actual}}{\text{680.5 g theor,}} \times 100 \, \% = \textbf{56.1 \%}\\\\\text{The percent yield is $\large \boxed{\mathbf{56.1 \, \% }}$}[/tex]
