Given the reaction below, answer the following questions. ( CH4(g) + S(g) → CS2 (g) + H2S(g) A.Balance the equation above. B.ls the sulfur oxidized or reduced in the reaction? C.Given 3.841x1025 molecules of CH4, how many moles is this? Show charges. D.Given 21.3 grams of Sulfur gas, how many moles of H2S(g) could be formed?

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answer A through D please

### Reaction Analysis and Stoichiometry

Given the reaction below, answer the following questions:

\[ \text{CH}_4(g) + \text{S}(g) \rightarrow \text{CS}_2(g) + \text{H}_2\text{S}(g) \]

#### Questions:
**A.** Balance the equation above.

**B.** Is the sulfur oxidized or reduced in the reaction? **Show charges.**

**C.** Given \( 3.841 \times 10^{25} \) molecules of \( \text{CH}_4 \), how many moles is this?

**D.** Given 21.3 grams of Sulfur gas, how many moles of \( \text{H}_2\text{S}(g) \) could be formed?

#### Solutions:

**A.** To balance the equation, one must make sure that the number of each type of atom on the left side of the equation is equal to the number on the right side.

\[ \text{CH}_4(g) + 4\text{S}(g) \rightarrow \text{CS}_2(g) + 2\text{H}_2\text{S}(g) \]

**B.** Determine whether sulfur is oxidized or reduced:

In \(\text{S}_8\) (solid sulfur), the oxidation state of sulfur is 0. In \(\text{CS}_2\) and \(\text{H}_2\text{S}\), the oxidation states of sulfur are as follows:
- In \(\text{CS}_2\): Sulfur has an oxidation state of -2.
- In \(\text{H}_2\text{S}\): Each sulfur atom has an oxidation state of -2.

Since sulfur goes from an oxidation state of 0 to an oxidation state of -2, it is **reduced**.

**C.** To convert molecules of \( \text{CH}_4 \) to moles:

Use Avogadro’s number \( (6.022 \times 10^{23}) \):

\[ \text{Number of moles} = \frac{\text{Number of molecules}}{\text{Avogadro's number}} \]

\[ = \frac{3.841 \times 10^{25}}{6.022 \times 10^{23}} \approx 63
Transcribed Image Text:### Reaction Analysis and Stoichiometry Given the reaction below, answer the following questions: \[ \text{CH}_4(g) + \text{S}(g) \rightarrow \text{CS}_2(g) + \text{H}_2\text{S}(g) \] #### Questions: **A.** Balance the equation above. **B.** Is the sulfur oxidized or reduced in the reaction? **Show charges.** **C.** Given \( 3.841 \times 10^{25} \) molecules of \( \text{CH}_4 \), how many moles is this? **D.** Given 21.3 grams of Sulfur gas, how many moles of \( \text{H}_2\text{S}(g) \) could be formed? #### Solutions: **A.** To balance the equation, one must make sure that the number of each type of atom on the left side of the equation is equal to the number on the right side. \[ \text{CH}_4(g) + 4\text{S}(g) \rightarrow \text{CS}_2(g) + 2\text{H}_2\text{S}(g) \] **B.** Determine whether sulfur is oxidized or reduced: In \(\text{S}_8\) (solid sulfur), the oxidation state of sulfur is 0. In \(\text{CS}_2\) and \(\text{H}_2\text{S}\), the oxidation states of sulfur are as follows: - In \(\text{CS}_2\): Sulfur has an oxidation state of -2. - In \(\text{H}_2\text{S}\): Each sulfur atom has an oxidation state of -2. Since sulfur goes from an oxidation state of 0 to an oxidation state of -2, it is **reduced**. **C.** To convert molecules of \( \text{CH}_4 \) to moles: Use Avogadro’s number \( (6.022 \times 10^{23}) \): \[ \text{Number of moles} = \frac{\text{Number of molecules}}{\text{Avogadro's number}} \] \[ = \frac{3.841 \times 10^{25}}{6.022 \times 10^{23}} \approx 63
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