Problem 11 - Chemistry and Chemical Reactivity

Question

Iron metal reacts with oxygen to give iron(III) oxide, \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) (a) Write a balanced equation for the reaction. (b) If an ordinary iron nail (assumed to be pure iron) has a mass of \(2.68 \mathrm{g},\) what mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) in grams, is produced if the nail is converted completely to the oxide? (c) What mass of \(\mathrm{O}_{2}\), in grams, is required for the reaction?

Step-by-Step Solution

Answer

(a) Balanced equation: \(4Fe + 3O_2 \rightarrow 2Fe_2O_3\). (b) Mass of \(Fe_2O_3\) produced: 3.83 g. (c) Mass of \(O_2\) required: 1.15 g.

Step 1: Write the Unbalanced Equation

First, identify the reactants and products of the reaction. Iron (Fe) reacts with oxygen (O_2) to form iron(III) oxide (Fe_2O_3). The unbalanced equation for the reaction is:\[ Fe + O_2 \rightarrow Fe_2O_3 \]

Step 2: Balance the Chemical Equation

Next, balance the chemical equation. Start by balancing the iron (Fe) atoms. Since there are two iron atoms in Fe_2O_3, you need two iron atoms as reactants:\[ 4Fe + O_2 \rightarrow 2Fe_2O_3 \] Now, balance the oxygen (O) atoms. There are three oxygen molecules in Fe_2O_3, so you need a total of six oxygen atoms as reactants:\[ 4Fe + 3O_2 \rightarrow 2Fe_2O_3 \]

Step 3: Calculate Moles of Iron

Calculate the moles of iron in the iron nail. Use the molar mass of iron, which is approximately 55.85 \, \text{g/mol}:\[ \text{Moles of } Fe = \frac{\text{mass of } Fe}{\text{molar mass of } Fe} = \frac{2.68 \, \text{g}}{55.85 \, \text{g/mol}} \approx 0.048 \, \text{mol} \]

Step 4: Calculate Moles of Iron(III) Oxide

Using the stoichiometry of the balanced equation, convert moles of Fe to moles of Fe_2O_3. The ratio is 4:2 (2:1):\[ \text{Moles of } Fe_2O_3 = \frac{0.048 \, \text{mol Fe}}{2} \approx 0.024 \, \text{mol Fe_2O_3} \]

Step 5: Calculate Mass of Iron(III) Oxide

Convert the moles of Fe_2O_3 to grams using its molar mass (159.7 \, \text{g/mol}):\[ \text{Mass of } Fe_2O_3 = 0.024 \, \text{mol} \times 159.7 \, \text{g/mol} = 3.8328 \, \text{g} \]

Step 6: Calculate Moles and Mass of Oxygen

Use the stoichiometry of the balanced equation to find the moles of O_2 required. For every 4 moles of Fe, 3 moles of O_2 are needed:\[ \text{Moles of } O_2 = \frac{0.048 \, \text{mol Fe} \times 3}{4} = 0.036 \, \text{mol O_2} \]Convert these moles to grams using the molar mass of O_2, which is 32 \, \text{g/mol}:\[ \text{Mass of } O_2 = 0.036 \, \text{mol} \times 32 \, \text{g/mol} = 1.152 \, \text{g} \]

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is an essential skill in chemistry that ensures the law of conservation of mass is upheld. This law states that matter cannot be created or destroyed in a chemical reaction; thus, the mass of the reactants must equal the mass of the products.

When we balance equations, each element must have the same number of atoms on both sides of the equation. Let's take the example of iron reacting with oxygen to form iron(III) oxide. Initially, we write the unbalanced reaction as:\[ \mathrm{Fe} + \mathrm{O}_2 \rightarrow \mathrm{Fe}_2\mathrm{O}_3 \]

In iron(III) oxide, there are two iron (Fe) atoms and three oxygen atoms. To balance the iron atoms, we place a coefficient of 4 in front of Fe as a reactant, resulting in 4 iron atoms on both sides:\[ 4\mathrm{Fe} + \mathrm{O}_2 \rightarrow 2\mathrm{Fe}_2\mathrm{O}_3 \]

Next, balance the oxygen atoms. We have six oxygen atoms needed in the products, which means we require three O₂ molecules as the reactant to provide 6 oxygen atoms, thereby completing the balanced equation:\[ 4\mathrm{Fe} + 3\mathrm{O}_2 \rightarrow 2\mathrm{Fe}_2\mathrm{O}_3 \]The equation is now balanced as both sides have equal numbers of Fe and O atoms.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction.

In a balanced chemical equation like \( 4\mathrm{Fe} + 3\mathrm{O}_2 \rightarrow 2\mathrm{Fe}_2\mathrm{O}_3 \), stoichiometry allows us to calculate the amount of products that will form given the number of reactants.

For example, if you have 0.048 moles of iron (Fe), the stoichiometric coefficients in the balanced equation tell us that 4 moles of Fe produce 2 moles of iron(III) oxide (\( \mathrm{Fe}_2\mathrm{O}_3 \)). Therefore, we can calculate the moles of \( \mathrm{Fe}_2\mathrm{O}_3 \):

• Given 0.048 moles of Fe, \( \frac{0.048 \text{ mol Fe}}{4} \times 2 = 0.024 \text{ mol } \mathrm{Fe}_2\mathrm{O}_3 \)

Similarly, we use the equation's stoichiometry to find how many moles of \( \mathrm{O}_2 \) are required:

• \( \frac{0.048 \text{ mol Fe} \times 3}{4} = 0.036 \text{ mol } \mathrm{O}_2 \)

Using stoichiometry, we efficiently convert between amounts of reactants and products.

Molar Mass Calculations

Knowing how to calculate molar mass is crucial when dealing with chemical reactions. Molar mass is the mass of one mole of a substance, and it lets us convert between grams and moles.

In our example, we are converting iron nails completely to iron(III) oxide. First, we need the molar mass of iron (Fe), which is approximately 55.85 \( \text{g/mol} \). Given the nail's mass at 2.68 g, the moles of Fe are calculated as:

• \( \text{Moles of } Fe = \frac{2.68 \text{ g}}{55.85 \text{ g/mol}} \approx 0.048 \text{ mol} \)

Next, we calculate the mass of iron(III) oxide with its molar mass (159.7 \( \text{g/mol} \)):

• Using 0.024 mol from stoichiometry, \( 0.024 \text{ mol} \times 159.7 \text{ g/mol} = 3.8328 \text{ g } \mathrm{Fe}_2\mathrm{O}_3 \)

Molar mass of oxygen \( (\mathrm{O}_2) \) is 32 \( \text{g/mol} \), which we use to find the needed grams for a complete reaction.

• \( 0.036 \text{ mol} \times 32 \text{ g/mol} = 1.152 \text{ g } \mathrm{O}_2 \)

Calculating molar mass and applying it to stoichiometry ensures precise predictions of reaction outcomes.