Problem 67 - Chemistry The Molecular Science

Question

In making iron from iron ore, this reaction occurs. $$\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{~g}) $$ (a) Calculate the mass of iron (in grams) that can be obtained from \(1.00 \mathrm{~kg}\) iron(III) oxide. (b) Calculate the mass of CO required.

Step-by-Step Solution

Answer

Approximately 699.1 g of iron and 526.1 g of CO are required.

Step 1: Write the Balanced Chemical Equation

The reaction is already provided balanced: \( \mathrm{Fe}_{2} \mathrm{O}_{3} + 3 \mathrm{CO} \rightarrow 2 \mathrm{Fe} + 3 \mathrm{CO}_{2} \). We will use this balanced equation to find the stoichiometric relationships for mass calculations.

Step 2: Calculate Moles of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \)

First, find the molar mass of iron(III) oxide, \( \mathrm{Fe}_{2} \mathrm{O}_{3} \). The molar mass is \( 2(55.85) + 3(16.00) = 159.70 \) g/mol. Convert 1.00 kg of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \) to grams: \( 1.00 \text{ kg} = 1000 \text{ g} \). Then, calculate moles of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \): \( \frac{1000 \text{ g}}{159.70 \text{ g/mol}} \approx 6.26 \text{ mol} \).

Step 3: Use Stoichiometry to Find Moles of \( \mathrm{Fe} \)

From the balanced equation, 1 mole of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \) produces 2 moles of \( \mathrm{Fe} \). Thus, \( 6.26 \text{ mol} \, \mathrm{Fe}_{2} \mathrm{O}_{3} \) produces \( 6.26 \times 2 = 12.52 \text{ mol} \) of \( \mathrm{Fe} \).

Step 4: Convert Moles of \( \mathrm{Fe} \) to Mass

Find the molar mass of \( \mathrm{Fe} \): 55.85 g/mol. Then calculate the mass of \( \mathrm{Fe} \): \( 12.52 \text{ mol} \times 55.85 \text{ g/mol} \approx 699.1 \text{ g} \). Therefore, approximately 699.1 g of iron can be obtained.

Step 5: Calculate Moles of \( \mathrm{CO} \) Required

For every mole of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \), 3 moles of \( \mathrm{CO} \) are required according to the balanced equation. Therefore, \( 6.26 \text{ mol} \times 3 = 18.78 \text{ mol} \) of \( \mathrm{CO} \) is needed.

Step 6: Convert Moles of \( \mathrm{CO} \) to Mass

Calculate the molar mass of \( \mathrm{CO} \): 12.01 (C) + 16.00 (O) = 28.01 g/mol. Thus, the mass of \( \mathrm{CO} \): \( 18.78 \text{ mol} \times 28.01 \text{ g/mol} \approx 526.1 \text{ g} \). Therefore, about 526.1 g of CO is required.

Key Concepts

Stoichiometry

In chemistry, stoichiometry is a powerful tool that helps us calculate the quantities of reactants and products in a chemical reaction. It is like a recipe that tells us how much of each ingredient is needed and what we get in return. When dealing with stoichiometry, the balanced chemical equation is our starting point as it provides the ratios in which substances react and produce. In the case of iron metallurgy, the reaction between iron(III) oxide, \( \mathrm{Fe}_{2}\mathrm{O}_{3} \), and carbon monoxide, \( \mathrm{CO} \), to produce iron \( \mathrm{Fe} \), and carbon dioxide \( \mathrm{CO}_{2} \), is represented by:

• \( \mathrm{Fe}_{2} \mathrm{O}_{3} + 3 \mathrm{CO} \rightarrow 2 \mathrm{Fe} + 3 \mathrm{CO}_{2} \)

This equation tells us that one mole of iron(III) oxide will react with three moles of carbon monoxide to produce two moles of iron and three moles of carbon dioxide. These relationships can then be used to determine the quantities of materials needed or produced in a given reaction scenario.

Chemical Equations

Chemical equations are the language of chemistry. They succinctly convey information about what happens during a chemical reaction, showing the reactants on the left side and the products on the right side of the arrow. Balanced chemical equations ensure that the same number of each type of atom appears on both sides of the equation, following the law of conservation of mass. For our reaction, \( \mathrm{Fe}_{2} \mathrm{O}_{3} + 3 \mathrm{CO} \rightarrow 2 \mathrm{Fe} + 3 \mathrm{CO}_{2} \), it is already balanced:

• There are 2 iron atoms, 3 carbon atoms, and 6 oxygen atoms on each side of the equation.
• This tells us that the amount of each atom is conserved throughout the reaction.

Understanding how to balance equations is a fundamental skill for solving stoichiometry problems, as it provides the necessary mole ratios for converting between moles of reactants and products.

Molar Mass

Molar mass is critical for converting between moles and grams of a substance. It is the mass of one mole of a substance and is given in grams per mole (g/mol). To calculate molar mass, you sum up the atomic masses of all the atoms in a molecule. For iron(III) oxide, \( \mathrm{Fe}_{2} \mathrm{O}_{3} \):

• Each \( \mathrm{Fe} \) atom has an atomic mass of 55.85 amu.
• Each \( \mathrm{O} \) atom has an atomic mass of 16.00 amu.
• Molar mass of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \) = \( 2 \times 55.85 + 3 \times 16.00 = 159.70 \) g/mol.

Knowing the molar mass allows chemists to calculate how much mass a certain number of moles of a compound will have, making it indispensable in both laboratory and industrial chemistry contexts.

Mass-to-Mole Conversion

Converting between mass and moles is a frequent necessity in chemistry calculations. The conversion relies on the relationship defined by the molar mass of a substance. The general formula for this conversion is: \(\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\). For example, to find how many moles are in 1 kg (or 1000 g) of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \), we use its molar mass of 159.70 g/mol:

• \( \frac{1000 \text{ g}}{159.70 \text{ g/mol}} \approx 6.26 \text{ mol} \).

Determining moles from mass is essential for then using stoichiometry to find out how much of each reactant is required or how much of a product you can expect from a reaction.