Problem 10 - Chemistry and Chemical Reactivity

Question

The balanced equation for a reaction in the process of reducing iron ore to the metal is $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g}) $$ (a) What is the maximum mass of iron, in grams, that can be obtained from \(454 \mathrm{g}(1.00 \mathrm{lb})\) of iron(III) oxide? (b) What mass of \(\mathrm{CO}\) is required to react with \(454 \mathrm{g}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3} ?\)

Step-by-Step Solution

Answer

317.43 g of iron can be obtained. 238.88 g of CO is required.

Step 1: Analyze the Chemical Equation

The balanced equation is given as \( \text{Fe}_{2}\text{O}_{3(s)} + 3 \text{CO}_{(g)} \rightarrow 2 \text{Fe}_{(s)} + 3 \text{CO}_{2(g)} \). This indicates that 1 mole of \( \text{Fe}_{2}\text{O}_3 \) reacts with 3 moles of \( \text{CO} \) to produce 2 moles of \( \text{Fe} \) and 3 moles of \( \text{CO}_2 \).

Step 2: Calculate Molar Masses

Calculate the molar masses of the compounds involved. The molar mass of \( \text{Fe}_{2}\text{O}_3 \) is \(2 \times 55.85 \text{ g/mol} + 3 \times 16.00 \text{ g/mol} = 159.70 \text{ g/mol}\). Similarly, the molar mass of \( \text{CO} \) is \(12.01 \text{ g/mol} + 16.00 \text{ g/mol} = 28.01 \text{ g/mol}\), and Fe is \(55.85 \text{ g/mol}\).

Step 3: Calculate Moles of Iron(III) Oxide

Using the mass given, calculate the moles of \( \text{Fe}_{2}\text{O}_3 \): \[ \text{Moles of Fe}_{2}\text{O}_3 = \frac{454 \text{ g}}{159.70 \text{ g/mol}} = 2.843 \text{ moles}\]

Step 4: Determine Moles of Iron Produced

From the balanced equation, 1 mole of \( \text{Fe}_{2}\text{O}_3 \) yields 2 moles of \( \text{Fe} \). Therefore, the moles of iron produced are:\[ 2 \times 2.843 = 5.686 \text{ moles of Fe}\]

Step 5: Calculate the Maximum Mass of Iron

Convert the moles of iron into mass using its molar mass:\[ \text{Mass of Fe} = 5.686 \text{ moles} \times 55.85 \text{ g/mol} = 317.43 \text{ g} \] Thus, the maximum mass of iron that can be obtained is 317.43 grams.

Step 6: Calculate Moles of Carbon Monoxide Needed

Using the balanced equation, 1 mole of \( \text{Fe}_{2}\text{O}_3 \) reacts with 3 moles of \( \text{CO} \). Therefore, the moles of \( \text{CO} \) needed are:\[ 3 \times 2.843 = 8.529 \text{ moles of CO}\]

Step 7: Determine Mass of Carbon Monoxide Required

Convert the moles of carbon monoxide into mass:\[ \text{Mass of CO} = 8.529 \text{ moles} \times 28.01 \text{ g/mol} = 238.88 \text{ g}\] Thus, the mass of carbon monoxide required is 238.88 grams.

Key Concepts

Chemical Reactions

Chemical reactions involve transforming reactants into products. In our exercise, we're looking at the reduction of iron(III) oxide to produce iron. This is shown by the reaction: \( \text{Fe}_{2}\text{O}_3(\text{s}) + 3 \text{CO}(\text{g}) \longrightarrow 2 \text{Fe}(\text{s}) + 3 \text{CO}_2(\text{g}) \). Here, iron(III) oxide reacts with carbon monoxide. The molecules rearrange, forming iron and carbon dioxide.

Each reactant and product has a different role:

• \(\text{Fe}_{2}\text{O}_3\): This is the iron compound we want to reduce, acting as the oxidizing agent.
• \(\text{CO}\): Serves as the reducing agent, helping convert \(\text{Fe}_{2}\text{O}_3\) into \(\text{Fe}\).
• \(\text{Fe}\): Our desired metal product, extracted from the ore.
• \(\text{CO}_2\): A by-product formed during the reaction.

This reaction showcases a fundamental process in metallurgy: extracting metals from ores.
Understanding the roles of each compound helps in predicting the reaction products and conditions needed.

Molar Mass Calculation

Molar mass is a crucial concept in stoichiometry. It gives the mass of one mole of a substance and allows us to convert between grams and moles. In the problem, we calculated the molar masses for each compound involved.

For \(\text{Fe}_{2}\text{O}_3\), each mole is made up of:

• 2 moles of iron \((2 \times 55.85 \text{ g/mol})\)
• 3 moles of oxygen \((3 \times 16.00 \text{ g/mol})\)

Adding these together gives a molar mass of \(159.70 \text{ g/mol}\).

For carbon monoxide, its molar mass is calculated by adding:

• 1 mole of carbon \((12.01 \text{ g/mol})\)
• 1 mole of oxygen \((16.00 \text{ g/mol})\)

This results in a molar mass of \(28.01 \text{ g/mol}\).

Understanding molar mass calculations aids in the conversion between mass and moles, a key step in solving stoichiometric problems.

Balancing Chemical Equations

Balancing chemical equations ensures the conservation of mass in a reaction. It means that the number of atoms of each element in the reactants must equal the number in the products. This principle is seen in our equation:

\( \text{Fe}_{2}\text{O}_3(\text{s}) + 3 \text{CO}(\text{g}) \longrightarrow 2 \text{Fe}(\text{s}) + 3 \text{CO}_2(\text{g}) \)

This equation is balanced, obeying the law of conservation of mass:

• 2 iron atoms on both sides.
• 3 oxygen and carbon atoms shared equally across the reactants and products.

Balancing allows us to accurately estimate the quantities of reactants needed and products formed in a reaction.

The coefficients (numbers before molecules) are key to balancing. They tell us the ratio in which substances react or are produced. Hence, balanced equations are fundamental in solving stoichiometry problems effectively.