QUESTION IMAGE
Question
sulfuryl chloride is a substance used to make chemicals for killing insects. it breaks down over time into sulfur dioxide and chlorine gas. consider the following reaction: some amount of sulfuryl chloride (so₂cl₂) breaks down to produce 2 molecules of sulfur dioxide (so₂) and 2 molecules of chlorine (cl₂). complete the table below. chemical element | number of atoms in the reaction o s cl during this reaction, how many molecules of sulfuryl chloride (so₂cl₂) react?
Part 1: Completing the table (for O, S, Cl atoms)
Step 1: Analyze Oxygen (O) atoms
- From \( SO_2 \): Each \( SO_2 \) has 2 O atoms, and there are 2 \( SO_2 \) molecules. So O atoms from \( SO_2 \): \( 2 \times 2 = 4 \).
- \( SO_2Cl_2 \) and \( Cl_2 \) have no O atoms. So total O atoms: \( 4 \).
Step 2: Analyze Sulfur (S) atoms
- From \( SO_2 \): Each \( SO_2 \) has 1 S atom, 2 \( SO_2 \) molecules. So S atoms from \( SO_2 \): \( 2 \times 1 = 2 \).
- From \( SO_2Cl_2 \): Let the number of \( SO_2Cl_2 \) be \( x \). Each \( SO_2Cl_2 \) has 1 S atom. By conservation of S, \( x \times 1 = 2 \), so \( x = 2 \) (we'll use this for S count too: \( 2 \times 1 = 2 \)).
Step 3: Analyze Chlorine (Cl) atoms
- From \( Cl_2 \): Each \( Cl_2 \) has 2 Cl atoms, 2 \( Cl_2 \) molecules. So Cl atoms from \( Cl_2 \): \( 2 \times 2 = 4 \).
- From \( SO_2Cl_2 \): Each \( SO_2Cl_2 \) has 2 Cl atoms, and we found \( x = 2 \) \( SO_2Cl_2 \) molecules. So Cl atoms from \( SO_2Cl_2 \): \( 2 \times 2 = 4 \) (matches conservation).
Part 2: Number of \( SO_2Cl_2 \) molecules
Step 1: Use conservation of S (or Cl, O)
- For S: Each \( SO_2Cl_2 \) has 1 S, each \( SO_2 \) has 1 S. There are 2 \( SO_2 \) molecules, so S atoms from \( SO_2 \) is 2. Thus, \( SO_2Cl_2 \) must provide 2 S atoms. Since each \( SO_2Cl_2 \) has 1 S, number of \( SO_2Cl_2 \) molecules is \( \frac{2}{1} = 2 \).
Table Completion:
| Chemical element | Number of atoms in the reaction |
|---|---|
| S | \( 2 \) |
| Cl | \( 4 \) |
Number of \( SO_2Cl_2 \) molecules:
The number of \( SO_2Cl_2 \) molecules that react is \( \boldsymbol{2} \).
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Part 1: Completing the table (for O, S, Cl atoms)
Step 1: Analyze Oxygen (O) atoms
- From \( SO_2 \): Each \( SO_2 \) has 2 O atoms, and there are 2 \( SO_2 \) molecules. So O atoms from \( SO_2 \): \( 2 \times 2 = 4 \).
- \( SO_2Cl_2 \) and \( Cl_2 \) have no O atoms. So total O atoms: \( 4 \).
Step 2: Analyze Sulfur (S) atoms
- From \( SO_2 \): Each \( SO_2 \) has 1 S atom, 2 \( SO_2 \) molecules. So S atoms from \( SO_2 \): \( 2 \times 1 = 2 \).
- From \( SO_2Cl_2 \): Let the number of \( SO_2Cl_2 \) be \( x \). Each \( SO_2Cl_2 \) has 1 S atom. By conservation of S, \( x \times 1 = 2 \), so \( x = 2 \) (we'll use this for S count too: \( 2 \times 1 = 2 \)).
Step 3: Analyze Chlorine (Cl) atoms
- From \( Cl_2 \): Each \( Cl_2 \) has 2 Cl atoms, 2 \( Cl_2 \) molecules. So Cl atoms from \( Cl_2 \): \( 2 \times 2 = 4 \).
- From \( SO_2Cl_2 \): Each \( SO_2Cl_2 \) has 2 Cl atoms, and we found \( x = 2 \) \( SO_2Cl_2 \) molecules. So Cl atoms from \( SO_2Cl_2 \): \( 2 \times 2 = 4 \) (matches conservation).
Part 2: Number of \( SO_2Cl_2 \) molecules
Step 1: Use conservation of S (or Cl, O)
- For S: Each \( SO_2Cl_2 \) has 1 S, each \( SO_2 \) has 1 S. There are 2 \( SO_2 \) molecules, so S atoms from \( SO_2 \) is 2. Thus, \( SO_2Cl_2 \) must provide 2 S atoms. Since each \( SO_2Cl_2 \) has 1 S, number of \( SO_2Cl_2 \) molecules is \( \frac{2}{1} = 2 \).
Table Completion:
| Chemical element | Number of atoms in the reaction |
|---|---|
| S | \( 2 \) |
| Cl | \( 4 \) |
Number of \( SO_2Cl_2 \) molecules:
The number of \( SO_2Cl_2 \) molecules that react is \( \boldsymbol{2} \).