Question

sulfur dioxide is a gas released when volcanoes erupt. it can react with oxygen gas to form sulfur trioxide, a chemical that can harm living things.

consider the following reaction:

12 molecules of sulfur dioxide (so₂) react with 6 molecules of oxygen (o₂) to produce some amount of sulfur trioxide (so₃).

complete the table below.

chemical element	number of atoms in the reaction
s

during this reaction, how many molecules of sulfur trioxide (so₃) are produced?

Explanation:

Part 1: Completing the table (Atoms Conservation)

For Oxygen (O) atoms:

• In $\ce{SO_2}$: Each $\ce{SO_2}$ has 2 O atoms. 12 molecules of $\ce{SO_2}$ have $12\times2 = 24$ O atoms.
• In $\ce{O_2}$: Each $\ce{O_2}$ has 2 O atoms. 6 molecules of $\ce{O_2}$ have $6\times2 = 12$ O atoms.
• Total O atoms in reactants: $24 + 12 = 36$. By conservation of mass, O atoms in products ( $\ce{SO_3}$) will also be 36.

For Sulfur (S) atoms:

• In $\ce{SO_2}$: Each $\ce{SO_2}$ has 1 S atom. 12 molecules of $\ce{SO_2}$ have $12\times1 = 12$ S atoms.
• In $\ce{O_2}$: There are 0 S atoms.
• Total S atoms in reactants: $12 + 0 = 12$. By conservation of mass, S atoms in products ( $\ce{SO_3}$) will also be 12.

Part 2: Molecules of $\ce{SO_3}$ produced

The balanced chemical equation for the reaction is:
$$\ce{2SO_2 + O_2 -> 2SO_3}$$

From the equation, 2 moles (or molecules) of $\ce{SO_2}$ react with 1 mole (or molecule) of $\ce{O_2}$ to produce 2 moles (or molecules) of $\ce{SO_3}$.

Given: 12 molecules of $\ce{SO_2}$ and 6 molecules of $\ce{O_2}$.

• Moles of $\ce{SO_2}$: 12, Moles of $\ce{O_2}$: 6.
• The ratio of $\ce{SO_2}$ to $\ce{O_2}$ in the reaction is 2:1. For 12 molecules of $\ce{SO_2}$, the required $\ce{O_2}$ is $\frac{12}{2} = 6$ molecules (which matches the given amount).

• From the equation, 2 molecules of $\ce{SO_2}$ produce 2 molecules of $\ce{SO_3}$. So, 12 molecules of $\ce{SO_2}$ will produce 12 molecules of $\ce{SO_3}$.

Table Completion:

Chemical element	Number of atoms in the reaction (Reactants: $\ce{SO_2}$ + $\ce{O_2}$; Products: $\ce{SO_3}$)
S	12 (from $\ce{SO_2}$) / 12 (in $\ce{SO_3}$)

Final Answer for Molecules of $\ce{SO_3}$:

\boxed{12}

Answer:

Part 1: Completing the table (Atoms Conservation)

For Oxygen (O) atoms:

• In $\ce{SO_2}$: Each $\ce{SO_2}$ has 2 O atoms. 12 molecules of $\ce{SO_2}$ have $12\times2 = 24$ O atoms.
• In $\ce{O_2}$: Each $\ce{O_2}$ has 2 O atoms. 6 molecules of $\ce{O_2}$ have $6\times2 = 12$ O atoms.
• Total O atoms in reactants: $24 + 12 = 36$. By conservation of mass, O atoms in products ( $\ce{SO_3}$) will also be 36.

For Sulfur (S) atoms:

• In $\ce{SO_2}$: Each $\ce{SO_2}$ has 1 S atom. 12 molecules of $\ce{SO_2}$ have $12\times1 = 12$ S atoms.
• In $\ce{O_2}$: There are 0 S atoms.
• Total S atoms in reactants: $12 + 0 = 12$. By conservation of mass, S atoms in products ( $\ce{SO_3}$) will also be 12.

Part 2: Molecules of $\ce{SO_3}$ produced

The balanced chemical equation for the reaction is:
$$\ce{2SO_2 + O_2 -> 2SO_3}$$

From the equation, 2 moles (or molecules) of $\ce{SO_2}$ react with 1 mole (or molecule) of $\ce{O_2}$ to produce 2 moles (or molecules) of $\ce{SO_3}$.

Given: 12 molecules of $\ce{SO_2}$ and 6 molecules of $\ce{O_2}$.

• Moles of $\ce{SO_2}$: 12, Moles of $\ce{O_2}$: 6.
• The ratio of $\ce{SO_2}$ to $\ce{O_2}$ in the reaction is 2:1. For 12 molecules of $\ce{SO_2}$, the required $\ce{O_2}$ is $\frac{12}{2} = 6$ molecules (which matches the given amount).

• From the equation, 2 molecules of $\ce{SO_2}$ produce 2 molecules of $\ce{SO_3}$. So, 12 molecules of $\ce{SO_2}$ will produce 12 molecules of $\ce{SO_3}$.

Table Completion:

Chemical element	Number of atoms in the reaction (Reactants: $\ce{SO_2}$ + $\ce{O_2}$; Products: $\ce{SO_3}$)
S	12 (from $\ce{SO_2}$) / 12 (in $\ce{SO_3}$)

Final Answer for Molecules of $\ce{SO_3}$:

\boxed{12}