Question

what is the percent yield of the reaction below when 544.5 g so2 and 160.0g o2 produce 382.0 g so3? 2SO2+O2+2SO3

Answer 1

`\large \boxed{56.1 \, \% }`

Step-by-step 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

`\text{Moles of SO}_(2) = \text{544.5 g} * \frac{\text{1 mol}}{\text{64.06 g }} = \text{8.500 mol}\\\\\text{Moles of O} _(2)= \text{160.0 g} * \frac{\text{1 mol}}{\text{32.00 g}} = \text{5.000 mol}`

(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

`\text{Moles of SO}_(3) = \text{8.500 mol SO}_(2) * \frac{\text{2 mol SO} _(3)}{\text{2 mol SO}_(2)}  = \text{8.500 mol SO}_(3)`

From O₂ :

The molar ratio of SO₂:O₂ is 2 mol O₂:1 mol O₂.

`\text{Moles of SO$_(3)$} = \text{5.000 mol O}_(2) * \frac{\text{2 mol SO$_(3)$}}{\text{1 mol O$_(2)$}} = \text{10.00 mol SO$_(3)$}`

The limiting reactant is SO₂ because it gives the smaller amount of SO₃.

(c) Calculate the theoretical yield of SO₃.

`\text{Theor. yield of SO}_(3) = \text{8.500 mol} * \frac{\text{80.06 g}}{\text{ 1 mol}} = \text{680.5 g}`

2. Calculate the percent yield of SO₃

`\text{\% yield} = \frac{\text{ 382.0 g actual}}{\text{680.5 g theor,}} * 100 \, \% =  \textbf{56.1 \%}\\\\\text{The percent yield is $\large \boxed{\mathbf{56.1 \, \% }}$}`