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?

Question

Gavyn

Chemistry

23 January, 23:54

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?

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Tyrell Frazier · Answer 1

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%.