**Chemical Reaction and Stoichiometry Analysis****Problem Statement:**Consider the balanced chemical reaction below. If 24.3 g of iron is produced when 50.0 g of iron(III) oxide and 50.0 g of carbon monoxide react, what is the percentage yield of the reaction?\[ \text{Fe}_2\text{O}_3(s) + 3 \text{CO}(g) \rightarrow 2 \text{Fe}(s) + 3 \text{CO}_2(g) \]**Progress Bar Illustration:**- A visual progress bar is presented, showing the user is on step 1 of 3.**Stoichiometry Table Setup Task:**Based on your knowledge of stoichiometry, set up the table below to determine the amounts of each reactant and product after the reaction goes to completion (assume 100% yield).**Table:**| | Fe₂O₃(s) | + | 3 CO(g) | → | 2 Fe(s) | + | 3 CO₂(g) ||-----------|----------|---|---------|---|--------|---|----------|| Before (mol) | [ ] | | [ ] | | [ ] | | [ ] || Change (mol) | [ ] | | [ ] | | [ ] | | [ ] || After (mol) | [ ] | | [ ] | | [ ] | | [ ] |*Note: The table provides blank spaces for entering the molar quantities before the reaction, the change during the reaction, and the quantities after the reaction for each substance involved.*This exercise requires an understanding of stoichiometric calculations to complete the table accurately. **Consider the balanced chemical reaction below. If 24.3 g of iron is produced when 50.0 g of iron(III) oxide and 50.0 g of carbon monoxide react, what is the percentage yield of the reaction?**\[ \text{Fe}_2\text{O}_3(s) + 3 \text{CO}(g) \rightarrow 2 \text{Fe}(s) + 3 \text{CO}_2(g) \]---- **Step Navigation**: A navigation bar shows the current step in a sequence, with step 2 highlighted. Steps 1 and 3 are also indicated, with options to move 'PREV' or 'NEXT.'- **Task**: Based on the setup from the previous step, determine the maximum number of grams of Fe that can be produced.\[ \text{mass}_{\text{Fe}} = \_\_\_\_\_ \text{g} \]---- **Options for Mass of Fe**: A series of clickable options are provided below the input box, displaying potential answers: 0, 17.5, 52.4, 47.2, 35.0, 99.7, 24.3, 50.0, 12.2.- **Reset Button**: There is a button labeled "RESET" to clear selections or calculations.

The amount of iron(III) oxide made to react with 50.0 g of carbon monoxide is 50.0 g. The amount of iron product is 24.3 g.The initial number of moles of iron(III) oxide is calculated by the formula, nFe2O3=mFe2O3MFe2O3 Where, mFe2O3 is the mass of iron(III) oxide. MFe2O3 is the molar mass of iron(III) oxide (159.69 g/mol). Substitute the mass and molar mass of iron(III) oxide in the above formula. nFe2O3=50.0 g159.69 g/mol=0.3131 molThe initial number of moles of carbon monoxide is calculated by the formula, nCO=mCOMCO Where, mCO is the mass of carbon monoxide. MCO is the molar mass of carbon monoxide (28.010 g/mol). Substitute the mass and molar mass of carbon monoxide in the above formula. nCO=50.0 g28.010 g/mol=1.785 molThe given chemical reaction shows that the molar ratio of iron(III) oxide and carbon monoxide is 1 : 3. The number of moles of carbon monoxide required to completely react with 0.3131 mol of iron(III) oxide is 0.3131 mol×3=0.9393 mol. Therefore, iron(III) is the limiting react. The given chemical reaction shows the molar ratio between iron(III) oxide and iron is 1 : 2. The amount of iron produced from 0.3131 mol of iron(III) oxide is calculated by the formula, mFe=2×nFe2O3×MFe Where, MFe is the molar mass of iron (55.845 g/mol). Substitute the number of moles of iron(III) oxide and the molar mass of iron in the above formula. mFe=2×0.3131 mol×55.845 g/mol=34.97 gThe percent yield is calculated by the formula, Percent yield=mProducedmFe×100 Where, mProduced is the amount of iron produced. Substitute the mass of iron produced and the mass of iron produced from 0.3131 mol of iron(III) oxide in the above formula. Percent yield=24.3 g34.97 g×100=69.5% Therefore, the percent yield of reaction is 69.5%.The initial number of moles of iron(III) oxide and carbon monoxide is 0.3131 mol and 1.785 mol. Iron(III) oxide is the limiting reactant. The ICE table (in terms of mol) is expressed as, Fe2O3 (s) + 3CO (g) → 2Fe (s) + 3CO2 (g) Initial (mol) 0.3131 1.785 mol 0 0 Change (mol) -0.3131 3×-0.3131 2×+0.3131 3×+0.3131 After (mol) 0 0.8457 0.6262 0.9393The maximum amount of iron that can be produced is calculated by the formula, mFe=nFe×MFe Where, MFe is the molar mass of iron (55.845 g/mol). nFe is the number of moles of iron. Substitute the number of moles of iron and the molar mass of iron in the above formula. mFe=0.6262 mol×55.845 g/mol=34.97 g≈35.0 gAnswer: The percent yield of the reaction is 69.5%. The number of moles of iron(III) oxide, carbon monoxide, iron and carbon dioxide formed after the reaction is 0 mol, 0.8457 mol, 0.6262 mol and 0.9393 mol, respectively. The maximum amount of iron that can be produced is 35.0 g.

Science

Chemistry

Consider the balanced chemical reaction below. If 24.3 g of iron is produced when 50.0 g of iron(III) oxide and 50.0 g of carbon monoxide react, what is the percentage yield of the reaction? Fe:O:(s) + 3 CO(g) → 2 Fe(s) + 3 CO:(g) 2 3 NEXT > Based on your knowledge of stoichiometry, set up the table below to determine the amounts of each reactant and product after the reaction goes to completion (assume 100% yield). Fe:O:(s) 3 CO(g) 2 Fe(s) 3 CO:(g) Before (mol) Change (mol) After (mol)

Consider the balanced chemical reaction below. If 24.3 g of iron is produced when 50.0 g of iron(III) oxide and 50.0 g of carbon monoxide react, what is the percentage yield of the reaction? Fe:O:(s) + 3 CO(g) → 2 Fe(s) + 3 CO:(g) 2 3 NEXT > Based on your knowledge of stoichiometry, set up the table below to determine the amounts of each reactant and product after the reaction goes to completion (assume 100% yield). Fe:O:(s) 3 CO(g) 2 Fe(s) 3 CO:(g) Before (mol) Change (mol) After (mol)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

Stoichiometry

Stoichiometry is a chemical division that evaluates the quantitative content of the chemical process with the help of the reacting molecules or products taking part in the reaction. The term “stoicheion” in “Stoichiometry”, is an old Greek term for "element" whereas “metron” means "measure".

Question

**Chemical Reaction and Stoichiometry Analysis**

**Problem Statement:**

Consider the balanced chemical reaction below. If 24.3 g of iron is produced when 50.0 g of iron(III) oxide and 50.0 g of carbon monoxide react, what is the percentage yield of the reaction?

\[ \text{Fe}_2\text{O}_3(s) + 3 \text{CO}(g) \rightarrow 2 \text{Fe}(s) + 3 \text{CO}_2(g) \]

**Progress Bar Illustration:**
- A visual progress bar is presented, showing the user is on step 1 of 3.

**Stoichiometry Table Setup Task:**

Based on your knowledge of stoichiometry, set up the table below to determine the amounts of each reactant and product after the reaction goes to completion (assume 100% yield).

**Table:**

| | Fe₂O₃(s) | + | 3 CO(g) | → | 2 Fe(s) | + | 3 CO₂(g) |
|-----------|----------|---|---------|---|--------|---|----------|
| Before (mol) | [ ] | | [ ] | | [ ] | | [ ] |
| Change (mol) | [ ] | | [ ] | | [ ] | | [ ] |
| After (mol) | [ ] | | [ ] | | [ ] | | [ ] |

*Note: The table provides blank spaces for entering the molar quantities before the reaction, the change during the reaction, and the quantities after the reaction for each substance involved.*

This exercise requires an understanding of stoichiometric calculations to complete the table accurately.

**Consider the balanced chemical reaction below. If 24.3 g of iron is produced when 50.0 g of iron(III) oxide and 50.0 g of carbon monoxide react, what is the percentage yield of the reaction?**

\[ \text{Fe}_2\text{O}_3(s) + 3 \text{CO}(g) \rightarrow 2 \text{Fe}(s) + 3 \text{CO}_2(g) \]

---

- **Step Navigation**: A navigation bar shows the current step in a sequence, with step 2 highlighted. Steps 1 and 3 are also indicated, with options to move 'PREV' or 'NEXT.'

- **Task**: Based on the setup from the previous step, determine the maximum number of grams of Fe that can be produced.

\[ \text{mass}_{\text{Fe}} = \_\_\_\_\_ \text{g} \]

---

- **Options for Mass of Fe**: A series of clickable options are provided below the input box, displaying potential answers: 0, 17.5, 52.4, 47.2, 35.0, 99.7, 24.3, 50.0, 12.2.

- **Reset Button**: There is a button labeled "RESET" to clear selections or calculations.

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