Z @ ▶ 12. How many grams of iron (III) oxide must you use? OA. 61.5g OB. 122 g OC. 0.777 g OD. 686 g 2 Text-to-Speech You are asked to make 43.0 grams of iron (Fe) from iron III oxide (Fe.Os) and carbon monoxide (CO) as shown in the chemical equation below. Fe₂O3 + 3C0→2Fe + 3C0₂ W S X # 15 3 # E + 1. $ 4 Q Search R D F 15 40 % 5 T CV 6 b "d+ VI B & Y 7 GH lip Lo 11 * N U 8 J 19 144 9 K M fro Dl O O tn bbl P : { [ ? I F 10 10 10 10 1

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### Chemistry Problem: Calculating the Mass of Iron(III) Oxide

**Question:**
You are asked to make 43.0 grams of iron (Fe) from iron III oxide (Fe₂O₃) and carbon monoxide (CO) as shown in the chemical equation below.

\[ \text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2 \]

How many grams of iron (III) oxide (Fe₂O₃) must you use?

**Options:**
- A. 61.5 g
- B. 122 g
- C. 0.777 g
- D. 686 g

---

**Explanation:**

To determine the amount of Fe₂O₃ needed to produce 43.0 grams of Fe, follow these steps:

1. **Calculate the Molar Masses:**

    - Molar mass of Fe (Iron) = 55.85 g/mol
    - Molar mass of Fe₂O₃ (Iron III Oxide):
      \[ \text{Fe}_2\text{O}_3 = (2 \times 55.85) + (3 \times 16.00) = 111.7 + 48.00 = 159.7 \text{ g/mol} \]

2. **Determine Moles of Fe Required:**

    \[ \text{Moles of Fe} = \frac{\text{Mass of Fe}}{\text{Molar Mass of Fe}} = \frac{43.0 \text{ g}}{55.85 \text{ g/mol}} \approx 0.770 \text{ mol Fe} \]

3. **Use Stoichiometry to Find Moles of Fe₂O₃ Needed:**
   
    According to the balanced equation, 1 mole of Fe₂O₃ produces 2 moles of Fe. Thus:
    \[ \text{Moles of Fe}_2\text{O}_3 = \frac{\text{Moles of Fe}}{2} = \frac{0.770 \text{ mol}}{2} \approx 0.385 \text{ mol Fe}_2\text{O}_3 \]

4. **Calculate the Mass of Fe₂O₃ Required:**

    \
Transcribed Image Text:### Chemistry Problem: Calculating the Mass of Iron(III) Oxide **Question:** You are asked to make 43.0 grams of iron (Fe) from iron III oxide (Fe₂O₃) and carbon monoxide (CO) as shown in the chemical equation below. \[ \text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2 \] How many grams of iron (III) oxide (Fe₂O₃) must you use? **Options:** - A. 61.5 g - B. 122 g - C. 0.777 g - D. 686 g --- **Explanation:** To determine the amount of Fe₂O₃ needed to produce 43.0 grams of Fe, follow these steps: 1. **Calculate the Molar Masses:** - Molar mass of Fe (Iron) = 55.85 g/mol - Molar mass of Fe₂O₃ (Iron III Oxide): \[ \text{Fe}_2\text{O}_3 = (2 \times 55.85) + (3 \times 16.00) = 111.7 + 48.00 = 159.7 \text{ g/mol} \] 2. **Determine Moles of Fe Required:** \[ \text{Moles of Fe} = \frac{\text{Mass of Fe}}{\text{Molar Mass of Fe}} = \frac{43.0 \text{ g}}{55.85 \text{ g/mol}} \approx 0.770 \text{ mol Fe} \] 3. **Use Stoichiometry to Find Moles of Fe₂O₃ Needed:** According to the balanced equation, 1 mole of Fe₂O₃ produces 2 moles of Fe. Thus: \[ \text{Moles of Fe}_2\text{O}_3 = \frac{\text{Moles of Fe}}{2} = \frac{0.770 \text{ mol}}{2} \approx 0.385 \text{ mol Fe}_2\text{O}_3 \] 4. **Calculate the Mass of Fe₂O₃ Required:** \
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