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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?

Respuesta :

Dejanras
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%.

Answer:

The percent yield of this reaction 93.75%.

Explanation:

[tex]2SO_2 (g) + O_2\rightarrow 2SO_3 (g)[/tex]

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

Moles of Sulfur dioxide =[tex]\frac{192 g}{64 g/mol}=3mol[/tex]

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

Then 3 mol of sulfur dioxide will give:

[tex]\frac{1}{1}\times 3 mol = 3mol[/tex] 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;

[tex]\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100[/tex]

The percent yield of this reaction:

[tex]\frac{225 g}{240 g}\times 100=93.75\%[/tex]

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