Question

its time for the unit exam. this is your opportunity to show how much you have if you dont feel ready, you may want to stop and clarify anything you do not und over any notes you have taken and review the key terms.
good luck!
note: to answer some of the questions, it may help to reference a periodic table
examine the statement.
sulfur reacts with molecular oxygen to produce sulfur trioxide.
which equation correctly represents the statement?
○ s + o₂ → so₃⁻²
○ 3s₂ + o₂ → 2s₃o
○ 2s + 3o₂ → 2so₃
○ s + 3o → so₃

Explanation:

Step1: Analyze the first option

The product in the first option \(S + O_{2}\to SO_{3}^{-2}\) is a sulfite ion (\(SO_{3}^{-2}\)), not sulfur trioxide (\(SO_{3}\)), so this is incorrect.

Step2: Analyze the second option

In the second option \(3S_{2}+O_{2}\to 2S_{3}O\), the formula of the product \(S_{3}O\) is not sulfur trioxide (\(SO_{3}\)), and also the oxygen in reactants is \(O_{2}\) (diatomic), but in the product it is in a wrong - combined form, so this is incorrect.

Step3: Analyze the third option

For the third option \(2S + 3O_{2}\to 2SO_{3}\), on the left - hand side:

• The number of sulfur atoms: \(2\) (from \(2S\))
• The number of oxygen atoms: \(3\times2 = 6\) (from \(3O_{2}\))

On the right - hand side:

• The number of sulfur atoms: \(2\) (from \(2SO_{3}\))
• The number of oxygen atoms: \(2\times3=6\) (from \(2SO_{3}\))

It follows the law of conservation of mass.

Step4: Analyze the fourth option

In the fourth option \(S + 3O\to SO_{3}\), oxygen in nature exists as \(O_{2}\) (diatomic) molecules, not as single \(O\) atoms in this reaction context, so this is incorrect.

Answer:

\(2S + 3O_{2}\to 2SO_{3}\)