Explanation:
# grams Sulfuric Acid formed from 22.3 g Sulfur Dioxide
[tex]23.2 \: g \: SO₂ \times \frac{1 \: mol \: SO₂)}{64.07g \: SO₂} \times \frac{1 \: mol \: H₂SO₃}{1 \: mol \: SO₂)} \times \frac{82.09 \: g \: H₂SO₃}{1 \: mol \:SO₂ } = 28.6 \: g \: H₂SO₃[/tex]
# grams Sulfuric Acid formed from 5.42 g Water
[tex]5.42 \: g \: SO₂ \frac{1 \: mol \: H₂O)}{18.02g \: H₂O} \times \frac{1 \: mol \: H₂SO₃}{1 \: mol \: H₂O)} \times \frac{82.09 \: g \: H₂SO₃}{1 \: mol \:H₂SO₃ } = 24.7 \: g \: H₂SO₃[/tex]
# Sulfur Dioxide used by reaction 5.42 g Water
[tex]5.42 \: g \: H₂O \times \frac{1 \: mol \: H₂O)}{18.02g \: H₂O} \times \frac{1 \: mol \: SO₂}{1 \: mol \: H₂O)} \times \frac{64.07 \: g \: SO₂}{1 \: mol \:SO₂ } = 19.27 \: g \: SO₂[/tex]
# of SO₂ in Excess
[tex] 23.2 \: g \: SO₂ - 19.27 \: g \: SO₂ = 3.0 \: g \: SO₂[/tex]
