Question

The reaction between solid sodium and iron(III) oxide is one in a series of reactions that inflates an automobile airbag: \(6 \mathrm{Na}(\mathrm{s})+\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s}) \rightarrow 3 \mathrm{Na}_{2} \mathrm{O}(\mathrm{s})+2 \mathrm{Fe}(\mathrm{s}) .\) If 100.0 \(\mathrm{g}\) of \(\mathrm{Na}\) and 100.0 \(\mathrm{g}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) are used in this reaction, determine the following. a. limiting reactant b. reactant in excess c. mass of solid iron produced d. mass of excess reactant that remains after the reaction is complete

Short answer

a. Limiting reactant: Fe₂O₃ b. Reactant in excess: Na c. Mass of solid iron produced: 70.0 g d. Mass of excess reactant that remains after the reaction is complete: 13.6 g

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the reaction between solid sodium and iron(III) oxide is provided in the exercise: \(6 Na(s) + Fe_{2}O_{3}(s) \rightarrow 3 Na_{2}O(s) + 2 Fe(s)\) We will use this balanced equation to determine the limiting reactant and the excess reactant.

Calculate the number of moles of reactants.

Given that 100.0 g of Na and 100.0 g of Fe₂O₃ are used in this reaction, we need to convert these amounts into moles using the molar masses of each substance. The molar mass of Na = 22.99 g/mol The molar mass of Fe₂O₃ = (2 × 55.85 g/mol) + (3 × 16.00 g/mol) = 159.69 g/mol Moles of Na = (100.0 g) / (22.99 g/mol) = 4.348 moles Moles of Fe₂O₃ = (100.0 g) / (159.69 g/mol) = 0.6263 moles

Determine the limiting reactant and excess reactant.

To determine which reactant is limiting, compare the mole ratio of the reactants with the stoichiometric ratio from the balanced equation. Divide the number of moles by the stoichiometric coefficient: For Na: \(4.348 / 6 = 0.7247\) For Fe₂O₃: \(0.6263 / 1 = 0.6263\) Since 0.6263 < 0.7247, Fe₂O₃ is the limiting reactant. Consequently, Na is the reactant in excess. a. Limiting reactant: Fe₂O₃ b. Reactant in excess: Na

Determine the mass of solid iron produced.

Now that we know the limiting reactant is Fe₂O₃, we can determine the mass of solid iron produced using stoichiometry. From the balanced equation, it's seen that 1 mole of Fe₂O₃ produces 2 moles of Fe. Therefore: Moles of Fe produced = (Moles of limiting reactant) × (Moles of Fe produced / Moles of Fe₂O₃) Moles of Fe produced = (0.6263) × (2 / 1) = 1.253 moles Now, we will convert the moles of Fe produced into grams using the molar mass of Fe (55.85 g/mol): Mass of Fe produced = (Moles of Fe produced) × (Molar mass of Fe) = 1.253 × 55.85 g/mol = 70.0 g c. Mass of solid iron produced: 70.0 g

Determine the mass of excess reactant that remains after the reaction is complete.

We need to determine how many moles of Na are consumed in the reaction. From the balanced equation, it's seen that 6 moles of Na react with 1 mole of Fe₂O₃. Therefore: Moles of Na consumed = (Moles of limiting reactant) × (Moles of Na consumed / Moles of Fe₂O₃) Moles of Na consumed = (0.6263) × (6 / 1) = 3.758 moles Now, determine the moles of Na left after the reaction: Moles of Na remaining = (Initial moles of Na) - (Moles of Na consumed) = 4.348 - 3.758 = 0.590 moles Finally, we will convert the moles of Na remaining into grams using the molar mass of Na (22.99 g/mol): Mass of Na remaining = (Moles of Na remaining) × (Molar mass of Na) = 0.590 × 22.99 g/mol = 13.6 g d. Mass of excess reactant that remains after the reaction is complete: 13.6 g

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that involves calculating the amounts of reactants and products in a chemical reaction. Its foundation lies in the balanced chemical equation, which tells us the exact proportion of substances involved. Consider the reaction between sodium (\(\text{Na}\)) and iron(III) oxide (\(\text{Fe}_2\text{O}_3\)). Using stoichiometry, we can predict the amount of product formed or the amount of reactants needed.
For instance, if we know the amount of one reactant, stoichiometry helps us find how much of another reactant is required or how much product will be formed. This concept is particularly useful when determining which reactant will run out first, known as the limiting reactant. Using mole ratios derived from the balanced chemical equation, stoichiometry guides us in these calculations.

• Calculate moles from grams (using molar mass).
• Use mole ratios to find relations between substances.
• Identify the limiting reactant using comparisons of available ratios.

Molar Mass

Molar mass is a crucial concept when dealing with chemical reactions, as it allows us to convert between the mass of a substance and the amount in moles. For any element or compound, it is the mass of one mole of that substance, typically expressed in grams per mole (\(\text{g/mol}\)).
Let's illustrate this with sodium (\(\text{Na}\)) and iron(III) oxide (\(\text{Fe}_2\text{O}_3\)):

• Sodium (Na): The molar mass is 22.99 g/mol. This means 22.99 grams of sodium is equal to one mole.
• Iron(III) oxide (Fe₂O₃): Calculated by adding twice the molar mass of iron (55.85 g/mol each) and thrice the molar mass of oxygen (16.00 g/mol each), totaling 159.69 g/mol.

These values are used to convert given masses of reactants or products in a chemical equation to moles, which is a necessary step in stoichiometry.

Chemical Reactions

Chemical reactions involve substances called reactants transforming into products. This transformation occurs at the molecular level where bonds break in reactants and new bonds form to create products. Analyzing reactions helps us understand nature's changes and predict outcomes.
In our example with sodium (\(\text{Na}\)) and iron(III) oxide (\(\text{Fe}_2\text{O}_3\)), these reactants react to produce sodium oxide (\(\text{Na}_2\text{O}\)) and iron (\(\text{Fe}\)).

• Reactants: Substances transformed during the reaction.
• Products: Substances formed as a result of the transformation.
• Reaction Details: Not all reactants are always entirely consumed. Some may remain as excess, making the understanding of limiting reactants vital.

Observing changes in mass, energy, and the properties of substances involved gives insight into the process of the reaction.

Balanced Chemical Equation

A balanced chemical equation reflects the conservation of mass by showing equal numbers of each atom type on both sides of the equation. This is vital as it depicts reality accurately, ensuring the law of conservation of mass is obeyed.
For instance, the equation \(6\text{Na(s)} + \text{Fe}_2\text{O}_3(\text{s}) \rightarrow 3 \text{Na}_2\text{O(s)} + 2 \text{Fe(s)}\) is balanced. It means:

• Sodium (Na) atoms: Six on the reactant side and six on the product side (as part of sodium oxide).
• Iron (Fe) atoms: Two on the reactant side, two on the product side.
• Oxygen (O) atoms: Three on each side, contained within iron(III) oxide and sodium oxide.

Balancing is accomplished by adjusting coefficients before each compound, not by altering subscripts within a formula. Correctly balancing a chemical equation is the starting point for any stoichiometric calculation, making it an essential skill in chemistry.