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Answered

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For the following reaction, 22.3 grams of sulfur dioxide are allowed to react with 5.42 grams of water.

sulfur dioxide (g) + water (l)→ sulfurous acid (H₂SO3) (g)

What is the maximum amount of sulfurous acid (H₂SO3) that can be formed? 24.7 grams

What is the FORMULA for the limiting reagent? H₂O

What amount of the excess reagent remains after the reaction is complete? 3.01 grams​

For The Following Reaction 223 Grams Of Sulfur Dioxide Are Allowed To React With 542 Grams Of Water Sulfur Dioxide G Water L Sulfurous Acid HSO3 G What Is The M class=

Sagot :

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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]

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