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Scorpion4ik [409]
3 years ago
12

3. In the formation of acid rain, sulfur dioxide reacts with oxygen and water in the air to form sulfuric acid. Write a balanced

chemical equation for the reaction. If 2.50 g sulfur dioxide react with excess oxygen gas and water, how many grams of sulfuric acid are produced
Chemistry
1 answer:
Andreyy893 years ago
4 0

Answer:

SO₂ + 0.5 O₂ + H₂O → H₂SO₄

3.83 g

Explanation:

In the formation of acid rain, sulfur dioxide reacts with oxygen and water in the air to form sulfuric acid. The balanced chemical equation is:

SO₂ + 0.5 O₂ + H₂O → H₂SO₄

The molar mass of SO₂ is 64.07 g/mol. The moles of SO₂ corresponding to 2.50 g are:

2.50 g × (1 mol/64.07 g) = 0.0390 mol

The molar ratio of SO₂ to H₂SO₄ is 1:1. The moles of H₂SO₄ formed are 0.0390 moles.

The molar mass of H₂SO₄ is 98.08 g/mol. The mass of H₂SO₄ is:

0.0390 mol × 98.08 g/mol = 3.83 g

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Answer: Option (d) is the correct answer.

Explanation:

Electronegativity value of hydrogen is 2.2.

Electronegativity value of chlorine is 3.16.

Electronegativity value of carbon is 2.55.

Electronegativity value of oxygen is 3.44.

Electronegativity value of nitrogen is 3.04.

Electronegativity value of sodium is 0.93.

Electronegativity value of iodine is 2.66.

Therefore, calculate the electronegativity difference between the bonded atoms as follows.

  • Electronegativity difference of HCl = Electronegativity value of chlorine - electronegativity value of hydrogen

                                                          = 3.16 - 2.2

                                                          = 0.96

  • Electronegativity difference of CO = Electronegativity value of oxygen - electronegativity value of carbon

                                                          =  3.44 - 2.55

                                                          = 0.89

  • Electronegativity difference of N_{2} = Electronegativity value of nitrogen - electronegativity value of nitrogen

                                                           = 3.04 - 3.04

                                                           = 0

  • Electronegativity difference of NaI = Electronegativity value of iodine - electronegativity value of sodium

                                                          = 2.66 - 0.93

                                                          = 1.73

So, we can see that highest electronegativity difference is 1.73 and it is shown by NaI molecule.

Thus, we can conclude that a group 1 alkali metal bonded to iodide, such as NaI has the greatest electronegativity difference between the bonded atoms.

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