Use the balanced equation and/or the periodic table information to write the conversion factor that belongs with the red arrow below. 6 PbO (s) + O2 (g) →2 Pb;O4 (s) 8 Grams Milliliters Grams of PbO of O2 of Pb;O4 Охудen 15.999 82 Pb Moles of Moles of Lead 207.2 Moles of PbO (s) O2 (g) Pb;O4 (s)

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Chapter10: Quantity Relationships In Chemical Reactions
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**Using Stoichiometry and the Periodic Table**

This educational segment explores the application of stoichiometry and periodic table information to determine conversion factors in chemical reactions. 

**Problem Statement:**
"Use the balanced equation and/or the periodic table information to write the conversion factor that belongs with the red arrow below."

**Balanced Equation:**
\[ 6 \text{PbO} \, (\text{s}) + \text{O}_2 \, (\text{g}) \rightarrow 2 \text{Pb}_3\text{O}_4 \, (\text{s}) \]

**Periodic Table Information:**
- **Oxygen (O)** - Atomic number: 8, Atomic mass: 15.999
- **Lead (Pb)** - Atomic number: 82, Atomic mass: 207.2

**Diagram Explanation:**
The diagram is a flowchart outlining the steps to convert between different quantities in the chemical reaction:

1. **Grams of PbO** are first converted into **moles of PbO (s)**.
2. The **moles of PbO (s)** are then converted into **moles of O2 (g)** via stoichiometric calculations based on the balanced chemical equation.
3. **Milliliters of O2** can be converted to **moles of O2 (g)** (if needed).
4. **Moles of O2 (g)** are converted to **moles of Pb3O4 (s)**.
5. Finally, the **moles of Pb3O4 (s)** are converted into **grams of Pb3O4**.

**Conversion Factor Question:**
In the diagram, focus is drawn to the conversion between **moles of O2 (g)** and **moles of Pb3O4 (s)** indicated by a red arrow.

**Key Step:**
Given the balanced equation:
\[ 6 \text{PbO} \, (\text{s}) + \text{O}_2 \, (\text{g}) \rightarrow 2 \text{Pb}_3\text{O}_4 \, (\text{s}) \]
The conversion factor between moles of \( \text{O}_2 \) and moles of \( \text{Pb}_3\text{O}_4 \) is derived from the stoichiometric coefficients:
\[ 1 \text{mole of O}_2 \rightarrow
Transcribed Image Text:**Using Stoichiometry and the Periodic Table** This educational segment explores the application of stoichiometry and periodic table information to determine conversion factors in chemical reactions. **Problem Statement:** "Use the balanced equation and/or the periodic table information to write the conversion factor that belongs with the red arrow below." **Balanced Equation:** \[ 6 \text{PbO} \, (\text{s}) + \text{O}_2 \, (\text{g}) \rightarrow 2 \text{Pb}_3\text{O}_4 \, (\text{s}) \] **Periodic Table Information:** - **Oxygen (O)** - Atomic number: 8, Atomic mass: 15.999 - **Lead (Pb)** - Atomic number: 82, Atomic mass: 207.2 **Diagram Explanation:** The diagram is a flowchart outlining the steps to convert between different quantities in the chemical reaction: 1. **Grams of PbO** are first converted into **moles of PbO (s)**. 2. The **moles of PbO (s)** are then converted into **moles of O2 (g)** via stoichiometric calculations based on the balanced chemical equation. 3. **Milliliters of O2** can be converted to **moles of O2 (g)** (if needed). 4. **Moles of O2 (g)** are converted to **moles of Pb3O4 (s)**. 5. Finally, the **moles of Pb3O4 (s)** are converted into **grams of Pb3O4**. **Conversion Factor Question:** In the diagram, focus is drawn to the conversion between **moles of O2 (g)** and **moles of Pb3O4 (s)** indicated by a red arrow. **Key Step:** Given the balanced equation: \[ 6 \text{PbO} \, (\text{s}) + \text{O}_2 \, (\text{g}) \rightarrow 2 \text{Pb}_3\text{O}_4 \, (\text{s}) \] The conversion factor between moles of \( \text{O}_2 \) and moles of \( \text{Pb}_3\text{O}_4 \) is derived from the stoichiometric coefficients: \[ 1 \text{mole of O}_2 \rightarrow
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