Question

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) together with other substances. Reaction of the ore with CO produces iron metal: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(\mathrm{s})+\mathrm{CO}_{2}(g) $$ (a) Balance this equation. (b) Calculate the number of grams of CO that can react with \(0.150 \mathrm{~kg}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) (c) Calculate the number of grams of Fe and the number of grams of \(\mathrm{CO}_{2}\) formed when \(0.150 \mathrm{~kg}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) reacts. (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

Short answer

The balanced equation for the reaction is \(Fe_{2}O_{3}(s) + 3CO(g) \longrightarrow 2Fe(s) + 3CO_{2}(g)\). For the reaction with 0.150 kg of \(Fe_{2}O_{3}\), 79.0 g of CO is needed. The resulting products are 105 g of Fe and 124 g of CO₂, which conforms to the law of conservation of mass.

Step-by-step solution

Count the elements on each side of the equation

Here, we will count the elements on both the reactants side and the products side of the equation. Reactants: Fe₂O₃ + CO || Products: Fe + CO₂ Reactants: Fe: 2 O: 3 C: 1 Products: Fe: 1 O: 2 C: 1

Balance the elements

We can see that Fe and O are not balanced. To balance the equation, we need to balance the Fe and O atoms on both sides. The balanced equation should be: \(Fe_{2}O_{3}(s) + 3CO(g) \longrightarrow 2Fe(s) + 3CO_{2}(g)\) #b. Calculate the mass of CO needed#

Convert the mass of Fe₂O₃ into moles

To calculate the mass of CO, first, we need to convert the mass of Fe₂O₃ into moles. We do this using the molar mass of Fe₂O₃. Given mass of Fe₂O₃ = 0.150 kg = 150 g Molar mass of Fe₂O₃ = (2 × 55.85 g/mol for Fe) + (3 × 16.00 g/mol for O) = 159.69 g/mol Moles of Fe₂O₃ = mass / molar mass = 150 g / 159.69 g/mol ≈ 0.940 mol

Calculate the moles of CO required

By using the stoichiometry in the balanced equation, we can find the moles of CO required for the reaction: 3 moles of CO are required for 1 mole of Fe₂O₃ Moles of CO = (3 moles CO/mole Fe₂O₃) × 0.940 mol Fe₂O₃ ≈ 2.820 mol

Convert moles of CO to mass

Now, we will convert moles of CO into grams by using the molar mass of CO. Molar mass of CO = (12.01 g/mol for C) + (16.00 g/mol for O) = 28.01 g/mol Mass of CO = moles × molar mass = 2.820 mol × 28.01 g/mol ≈ 79.0 g The mass of CO that can react with 0.150 kg of Fe₂O₃ is 79.0 g. #c. Calculate the grams of Fe and CO₂ formed#

Calculate the moles of Fe and CO₂ formed

From the balanced equation, we can determine the moles of Fe and CO₂ formed in the reaction: 2 moles of Fe and 3 moles of CO₂ are formed for 1 mole of Fe₂O₃ reacting Moles of Fe: (2 moles Fe/mole Fe₂O₃) × 0.940 mol Fe₂O₃ ≈ 1.880 mol Moles of CO₂: (3 moles CO₂/mole Fe₂O₃) × 0.940 mol Fe₂O₃ ≈ 2.820 mol

Calculate the mass of Fe and CO₂

Now, we will convert the moles of Fe and CO₂ into grams. Molar mass of Fe = 55.85 g/mol Molar mass of CO₂ = (12.01 g/mol for C) + (2 × 16.00 g/mol for O) = 44.01 g/mol Mass of Fe = moles × molar mass = 1.880 mol × 55.85 g/mol ≈ 105 g Mass of CO₂ = moles × molar mass = 2.820 mol × 44.01 g/mol ≈ 124 g The grams of Fe and CO₂ formed when 0.150 kg of Fe₂O₃ reacts are 105 g and 124 g, respectively. #d. Conservation of mass#

Calculate the mass of reactants and products

Mass of reactants = Mass of Fe₂O₃ + Mass of CO = 150 g + 79.0 g = 229 g Mass of products = Mass of Fe + Mass of CO₂ = 105 g + 124 g = 229 g

Check if the mass is conserved

The mass of reactants (229 g) is equal to the mass of products (229 g), and therefore, the calculations are consistent with the law of conservation of mass.

Key concepts

Chemical Equation Balancing

Understanding how to balance a chemical equation is crucial in the field of chemistry. It represents the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. To balance an equation, we ensure that the number of atoms for each element is the same on both the reactant and product sides.

In our exercise example, we balance the equation \[Fe_{2}O_{3}(s) + CO(g) \longrightarrow Fe(s) + CO_{2}(g)\] by adjusting the coefficients, which are the numbers placed before compounds in a reaction. The final balanced equation is \[Fe_{2}O_{3}(s) + 3CO(g) \longrightarrow 2Fe(s) + 3CO_{2}(g)\].

Here's a simplified process to balance equations:

• Identify the reactants and products.
• List the number of atoms of each element present in the unbalanced equation.
• Use coefficients to obtain the same number of atoms of each element on each side.
• Check your work to confirm that all elements are balanced and the smallest whole number coefficients are used.

Molar Mass Calculation

The molar mass is the weight of one mole of a substance, typically expressed in grams per mole (g/mol). It is calculated by summing up the atomic masses of each element in a compound multiplied by their respective number of atoms, as per the compound's chemical formula.

For instance, Iron (III) oxide \(Fe_{2}O_{3}\) has a molar mass calculated by \[(2 \times 55.85 \text{ g/mol for } Fe) + (3 \times 16.00 \text{ g/mol for } O) = 159.69 \text{ g/mol}\].

Accurate molar mass is necessary for converting between moles and grams, which allows chemists to measure out exact amounts of a substance.

Stoichiometric Calculations

Stoichiometry is like the recipe for a chemical reaction. It relates the amounts of reactants to products using the balanced chemical equation and is crucial for determining how much of each substance is used or created.

In the case of our iron ore reaction, we first convert mass to moles because stoichiometry deals with moles rather than mass. Then, using the balanced equation \[Fe_{2}O_{3}(s) + 3CO(g) \longrightarrow 2Fe(s) + 3CO_{2}(g)\], we find that 3 moles of CO are required to react with 1 mole of \(Fe_{2}O_{3}\). To complete stoichiometric calculations:

• Convert all given information into moles, if not already.
• Use the mole ratio from the balanced equation to determine the moles of other substances involved.
• Convert moles back to desired units.
• Use the law of conservation of mass to check calculations for possible errors.

Through stoichiometry, we can predict the outcomes of the chemical reactions quantitatively, making it a cornerstone of chemical and industrial processes.