d Iron metal reacts with chlorine gas giving iron(III) chloride. The balanced chemical equation for this reaction is: 2 Fe(s) + 3 Cl₂ (g) → 2 FeCl3 (s) If 32.0 g each of iron and chlorine are combined, what is the theoretical yield of iron(III) chloride? 9

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### Chemical Reactions and Stoichiometry: Example Problem

#### Problem Statement
Iron metal reacts with chlorine gas to produce iron(III) chloride. The balanced chemical equation for this reaction is:
\[ 2 \text{Fe}(s) + 3 \text{Cl}_2(g) \rightarrow 2 \text{FeCl}_3(s) \]

#### Question
If 32.0 g of iron and 32.0 g of chlorine are combined, what is the theoretical yield of iron(III) chloride?

\[ \boxed{\hspace{1cm}} \text{g} \]

#### Solution Steps
1. **Balanced Chemical Equation:** Given \(2 \text{Fe}(s) + 3 \text{Cl}_2(g) \rightarrow 2 \text{FeCl}_3(s)\), this indicates that 2 moles of iron react with 3 moles of chlorine gas to produce 2 moles of iron(III) chloride.

2. **Molar Mass Calculation:**
   - Molar mass of Fe (iron) = 55.85 g/mol
   - Molar mass of \( \text{Cl}_2 \) (chlorine gas) = 2 * 35.45 g/mol = 70.90 g/mol
   - Molar mass of \( \text{FeCl}_3 \) (iron(III) chloride) = 55.85 g/mol + 3 * 35.45 g/mol = 162.20 g/mol

3. **Determine Moles of Reactants:**
   - Moles of iron = \( \frac{32.0 \text{ g}}{55.85 \text{ g/mol}} \approx 0.573 \text{ moles}\)
   - Moles of chlorine = \( \frac{32.0 \text{ g}}{70.90 \text{ g/mol}} \approx 0.451 \text{ moles}\)

4. **Determine the Limiting Reactant:**
   - According to the balanced equation, 2 moles of Fe react with 3 moles of \( \text{Cl}_2 \).
   - Therefore, for 0.573 moles of Fe, we would need \( 0.573 \times \frac{3}{2} = 0.860 \text{ moles} \) of Cl\(_
Transcribed Image Text:### Chemical Reactions and Stoichiometry: Example Problem #### Problem Statement Iron metal reacts with chlorine gas to produce iron(III) chloride. The balanced chemical equation for this reaction is: \[ 2 \text{Fe}(s) + 3 \text{Cl}_2(g) \rightarrow 2 \text{FeCl}_3(s) \] #### Question If 32.0 g of iron and 32.0 g of chlorine are combined, what is the theoretical yield of iron(III) chloride? \[ \boxed{\hspace{1cm}} \text{g} \] #### Solution Steps 1. **Balanced Chemical Equation:** Given \(2 \text{Fe}(s) + 3 \text{Cl}_2(g) \rightarrow 2 \text{FeCl}_3(s)\), this indicates that 2 moles of iron react with 3 moles of chlorine gas to produce 2 moles of iron(III) chloride. 2. **Molar Mass Calculation:** - Molar mass of Fe (iron) = 55.85 g/mol - Molar mass of \( \text{Cl}_2 \) (chlorine gas) = 2 * 35.45 g/mol = 70.90 g/mol - Molar mass of \( \text{FeCl}_3 \) (iron(III) chloride) = 55.85 g/mol + 3 * 35.45 g/mol = 162.20 g/mol 3. **Determine Moles of Reactants:** - Moles of iron = \( \frac{32.0 \text{ g}}{55.85 \text{ g/mol}} \approx 0.573 \text{ moles}\) - Moles of chlorine = \( \frac{32.0 \text{ g}}{70.90 \text{ g/mol}} \approx 0.451 \text{ moles}\) 4. **Determine the Limiting Reactant:** - According to the balanced equation, 2 moles of Fe react with 3 moles of \( \text{Cl}_2 \). - Therefore, for 0.573 moles of Fe, we would need \( 0.573 \times \frac{3}{2} = 0.860 \text{ moles} \) of Cl\(_
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