Respuesta :

The mass of sulfurous acid produced when 245g of sulfur dioxide is reacted with water is 313.8 g

Stoichiometry

From the question, we are to determine the mass of sulfurous acid produced

First, we will write the balaced chemical equation for the reaction

H₂O + SO₂ → H₂SO₃

This means 1 mole of H₂O reacts with 1 mole of sulfur dioxide (SO₂) to produce 1 mole of sulfurous acid (H₂SO₃)

Now, we will determine the number of moles of sulfur dioxide present

Mass of sulfur dioxide present = 245 g

Molar mass of sulfur dioxide = 64.066 g/mol

Using the formula,

[tex]Number \ of \ moles = \frac{Mass}{Molar\ mass}[/tex]

Then,

Number of moles of SO₂ present = [tex]\frac{245}{64.066}[/tex]

Number of moles of SO₂ present = 3.824 moles

This is the number of moles of SO₂ present

From balanced chemical equation

1 mole of H₂O reacts with 1 mole of sulfur dioxide (SO₂) to produce 1 mole of sulfurous acid (H₂SO₃)

Then,

Water (H₂O) will react with the 3.824 moles of sulfur dioxide (SO₂) to produce 3.824 moles of sulfurous acid (H₂SO₃)

∴ The number of moles of sulfurous acid produced is 3.824 moles

Now, for the mass of sulfurous acid produced

Molar mass of sulfurous acid = 82.07 g/mole

Using the formula,

Mass = Number of moles × Molar mass

Mass of sulfurous acid produced = 3.824 × 82.07

Mass of sulfurous acid produced = 313.8 g

Hence, the mass of sulfurous acid produced is 313.8 g

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