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Sagot :
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
Learn more on Stoichiometry here: https://brainly.com/question/6061451
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