Gotenks99
Answered

Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Explore our Q&A platform to find in-depth answers from a wide range of experts in different fields. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

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 :

7142ja99

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]

Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.