Problem 23 - Chemistry Matter and Change

Question

The reaction between solid sodium and iron(llI) 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. \begin{equation} \begin{array}{l}{\text { a. limiting reactant }} \\ {\text { b. reactant in excess }} \\ {\text { c. mass of solid iron produced }} \\ {\text { d. mass of excess reactant that remains after the reaction is complete }}\end{array} \end{equation}

Step-by-Step Solution

Answer

a. The limiting reactant is Fe\(_2\)O\(_3\). b. The reactant in excess is Na. c. The mass of solid iron produced is 69.89 g. d. The mass of excess sodium that remains is 13.66 g.

Step 1: Calculate Moles of Sodium

First, we need to convert the mass of sodium to moles. Sodium (Na) has a molar mass of 22.99 g/mol. Using this data, we calculate the moles:\[\text{Moles of Na} = \frac{100.0 \text{ g}}{22.99 \text{ g/mol}} = 4.35 \text{ mol}\]

Step 2: Calculate Moles of Iron(III) Oxide

Next, convert the mass of iron(III) oxide to moles. The molar mass of \( \text{Fe}_2\text{O}_3 \) (iron(III) oxide) is 159.7 g/mol:\[\text{Moles of } \text{Fe}_2\text{O}_3 = \frac{100.0 \text{ g}}{159.7 \text{ g/mol}} = 0.626 \text{ mol}\]

Step 3: Determine Limiting Reactant

According to the balanced equation, 6 moles of Na react with 1 mole of \( \text{Fe}_2\text{O}_3 \). Convert available moles with given ratios:\[\text{Required moles of Na for } 0.626 \text{ mol } \text{Fe}_2\text{O}_3 = 6 \times 0.626 = 3.756 \text{ mol Na}\]Since 4.35 moles of Na are available and only 3.756 moles are required, Fe\(_2\)O\(_3\) is the limiting reactant.

Step 4: Identify Reactant in Excess

Since \( \text{Fe}_2\text{O}_3 \) is limiting, sodium (Na) is in excess.

Step 5: Calculate Mass of Produced Iron

Using the stoichiometry from the balanced equation, 1 mole of \( \text{Fe}_2\text{O}_3 \) produces 2 moles of Fe:\[\text{Moles of Fe produced} = 2 \times 0.626 = 1.252 \text{ mol Fe}\]The molar mass of iron (Fe) is 55.85 g/mol, so:\[\text{Mass of Fe} = 1.252 \times 55.85 = 69.89 \text{ g}\]

Step 6: Calculate Mass of Excess Sodium Remaining

Calculate the moles of Na actually used in reaction, which is 3.756 moles. Subtract the used amount from the initial moles to get the excess.\[\text{Excess moles of Na} = 4.35 - 3.756 = 0.594 \text{ mol} \]Convert back to grams using molar mass of Na:\[\text{Mass of excess Na} = 0.594 \times 22.99 = 13.66 \text{ g}\]

Key Concepts

Stoichiometry

Stoichiometry is an essential concept in chemistry that involves the calculation of reactants and products in chemical reactions. It forms the mathematical backbone of chemical equations. By using stoichiometry, we can determine how much of a reactant will be needed or how much of a product will be formed during a reaction.
In the given problem, stoichiometry allows us to use the balanced chemical equation:

• \[6 \text{Na} (\text{s}) + \text{Fe}_2\text{O}_3 (\text{s}) \rightarrow 3 \text{Na}_2\text{O} (\text{s}) + 2 \text{Fe} (\text{s})\]

to determine the proportions of each substance. By comparing available quantities of reactants to these proportions, we can identify the limiting reactant and predict the amount of product (iron) formed. This calculation is performed by understanding the mole ratios, which are derived from the coefficients in the balanced equation.
Understanding stoichiometry is crucial for predicting how a change in quantity of one substance will impact the rest of the reaction.

Molar Mass

Molar mass is the mass of one mole of a given substance and is expressed in grams per mole (g/mol). It acts like a conversion factor between the mass of a substance and the amount of substance in moles.
Take sodium (Na), for example, which has a molar mass of 22.99 g/mol. This means that one mole of sodium atoms weighs 22.99 grams. Similarly, iron(III) oxide (\(\text{Fe}_2\text{O}_3\)) has a molar mass of 159.7 g/mol. To calculate the number of moles for a given mass, we use the formula:

• \[\text{Moles} = \frac{\text{mass of substance}}{\text{molar mass}}\]

In our problem, this formula was used twice: once to convert 100 grams of sodium into moles, and again to convert 100 grams of iron(III) oxide into moles. Knowing these molar masses is a fundamental step in solving stoichiometry problems as it helps us compare the quantities of different substances involved.

Chemical Reaction

A chemical reaction is a process in which substances (reactants) undergo chemical changes, forming new substances (products). This process is mapped out using a chemical equation, which symbolizes the reactants and products along with their respective quantities.
In the reaction between sodium (Na) and iron(III) oxide (\(\text{Fe}_2\text{O}_3\)), sodium is oxidized, forming sodium oxide (\(\text{Na}_2\text{O}\)), while iron(III) oxide is reduced, producing iron metal (Fe). This reaction can be succinctly expressed with the balanced equation:

• \[6 \text{Na} (\text{s}) + \text{Fe}_2\text{O}_3 (\text{s}) \rightarrow 3 \text{Na}_2\text{O} (\text{s}) + 2 \text{Fe} (\text{s})\]

Each molecule or formula unit in the equation is preceded by a coefficient, which indicates the number of moles needed or produced. Through understanding and balancing chemical equations, we can predict how many moles of each substance are involved, guiding us in determining the limiting reactant and excess reactants in a chemical reaction.

Excess Reactant

In chemical reactions, the excess reactant is the substance that remains after the reaction has completed. It is the reactant that is not completely consumed when the reaction reaches completion due to having more present than required to react completely with the limiting reactant.
From our problem, iron(III) oxide (\(\text{Fe}_2\text{O}_3\)) is identified as the limiting reactant, which means sodium (Na) is in excess. This can be determined by comparing the calculated moles required from the balanced equation to those that are actually available.
Once we've established that sodium is in excess, we can calculate the remaining amount after the reaction is complete. The initial moles of sodium minus the moles that reacted gives us the moles of sodium left over, allowing us to convert this back into grams using sodium's molar mass. Thus, identifying the mass of the excess reactant provides insight into resource efficiency and helps to determine the cost-effectiveness of industrial chemical processes.