Question

In the manufacturing process of sulfuric acid, sulfur dioxide is reacted with oxygen to produce sulfur trioxide. Using the equation, 2SO2 (g) + O2 Imported Asset 2SO3 (g), if 192 g of sulfur dioxide is given the opportunity to react with an excess of oxygen to produce 225 g of sulfur trioxide, what is the percent yield of this reaction?

Answer 1

The percent yield of this reaction 93.75%.

Step-by-step explanation:

`2SO_2 (g) + O_2\rightarrow 2SO_3 (g)`

if 192 g of sulfur dioxide reacts with excessive of oxygen gas.

Moles of Sulfur dioxide =
`(192 g)/(64 g/mol)=3mol`

According to reaction 2 moles of sulfur dioxide gives 2 moles of sulfur trioxide.

Then 3 mol of sulfur dioxide will give:

`(1)/(1)* 3 mol = 3mol` of sulfur trioxide

Mass of 3 mole of sulfur trioxide = 3 mol × 80 g/mol = 240 g

Experimental yield of sulfur tiroxide = 240 g

Theoretical yield of sulfur trioxide = 225 g

Percentage yield;

`\frac{\text{Experimental yield}}{\text{theoretical yield}}* 100`

The percent yield of this reaction:

`(225 g)/(240 g)* 100=93.75\%`

Answer 2

Answer is: yield of this reaction is 93,66%.
Chemical reaction: 2SO₂ + O₂ → 2SO₃.
m(SO₂) = 192 g.
m(SO₃) = 225 g.
n(SO₂) = m(SO₂) ÷ M(SO₂).
n(SO₂) = 192 g ÷ 64 g/mol.
n(SO₂) = 3 mol.
n(SO₃) = m(SO₃) ÷ M(SO₃).
n(SO₃) = 225 g ÷ 80 g/mol.
n(SO₃) = 2,81 mol.
From chemical reaction: n(SO₂) : n(SO₃) = 2 : 2.
n(SO₃) = 3 mol.
Yield of reaction: 2,81 mol ÷ 3 mol · 100% = 93,66%.