Problem 45 - Chemistry A Molecular Approach

Question

Iron(III) oxide reacts with carbon monoxide according to the equation: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g) $$ A reaction mixture initially contains \(22.55 \mathrm{~g} \mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(14.78 \mathrm{~g}\) CO. Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant remains?

Step-by-Step Solution

Answer

2.915 g of CO remains unreacted.

Step 1: Calculate molar masses

First, calculate the molar masses of iron(III) oxide (\text{Fe}_2\text{O}_3) and carbon monoxide (CO). For \text{Fe}_2\text{O}_3, the molar mass is \(2 \times 55.85\) (iron) + \(3 \times 16.00\) (oxygen) = \(159.7\text{g/mol}\). For CO, the molar mass is \(12.01\) (carbon) + \(16.00\) (oxygen) = \(28.01\text{g/mol}\).

Step 2: Calculate moles of reactants

Using the molar masses, calculate the moles of reactants. For \(\text{Fe}_2\text{O}_3\), moles = \(\frac{22.55\text{ g}}{159.7\text{ g/mol}} = 0.1412\text{ mol}\). For CO, moles = \(\frac{14.78\text{ g}}{28.01\text{ g/mol}} = 0.5277\text{ mol}\).

Step 3: Determine the limiting reactant

Based on the stoichiometry of the reaction, 1 mole of \(\text{Fe}_2\text{O}_3\) reacts with 3 moles of CO. Compare the mole ratio of the available reactants with the stoichiometric ratio. \(0.1412\) mol \(\text{Fe}_2\text{O}_3\) would need \(0.1412 \times 3 = 0.4236\) mol CO, but there is \(0.5277\) mol CO available. Hence, \(\text{Fe}_2\text{O}_3\) is the limiting reactant and CO is in excess.

Step 4: Calculating the excess amount of CO

Since \(\text{Fe}_2\text{O}_3\) limits the reaction, calculate the amount of CO that remains unreacted: \(0.5277\text{ mol} - 0.4236\text{ mol} = 0.1041\text{ mol}\) of CO is left over.

Step 5: Calculating mass of excess CO

The mass of the excess CO is the number of moles multiplied by its molar mass: \(0.1041\text{ mol} \times 28.01\text{ g/mol} = 2.915\text{ g}\). This is the mass of CO that remains after the reaction is complete.

Key Concepts

Limiting Reactant

The concept of the limiting reactant is a fundamental aspect of stoichiometry in chemical reactions. It's pivotal to understand that in a chemical reaction, the reactant that is used up first limits the amount of product that can be formed. Think of it as a bottleneck in production: no matter how much of the other reactants are available, if you run out of one, you can't make any more product.

Let's consider a simple analogy: If you're making sandwiches and you have 10 slices of bread and only 1 slice of cheese, your cheese is your limiting reactant. You can only make one sandwich, and those extra slices of bread won't change that. Similarly, in the provided exercise, iron(III) oxide reacts with carbon monoxide, but because of the amounts we started with, the iron(III) oxide will run out first, dictating the maximum amount of iron that can be produced.

It's important when finding the limiting reactant to compare mole ratios, not just the mass of the reactants. The steps outlined in the solution demonstrate exactly this process. First, we figure out how many moles we have of each reactant, then determine which reactant has the mole ratio that is lower than what is required by the reaction stoichiometry. This is a critical skill in chemistry, as it allows for precise predictions about the outcomes of reactions.

Molar Mass Calculation

Understanding molar mass calculation is another cornerstone of stoichiometry and is essential for converting between grams and moles -- which is like translating between two languages in chemistry.

The molar mass represents the mass of one mole of a substance (usually in grams per mole, g/mol) and is calculated by summing up the atomic masses of all the atoms in a molecule. These atomic masses can be found on the periodic table and usually take into account the average mass of all isotopes of an element.

In the exercise, for example, iron(III) oxide has a molar mass that is determined by adding the atomic masses of two iron atoms and three oxygen atoms. Knowing the molar mass allows us to calculate how many moles of each reactant we have. Without correctly calculating molar masses, none of the subsequent steps in the process would yield the right answer. It's much like knowing the conversion rate between currencies when trying to budget travel expenses in a foreign country. Molar mass gives us the 'exchange rate' for going from grams to moles and back, enabling us to work with quantities in a meaningful way within chemical equations.

Chemical Reaction Balance

The balance of a chemical reaction is the bread and butter of stoichiometry. Just like in an algebraic equation, what goes into a reaction (reactants) must equal what comes out (products). However, instead of numbers, we balance the type and number of atoms.

A balanced chemical reaction ensures that the law of conservation of mass is adhered to, meaning matter is neither created nor destroyed during the course of a chemical reaction. When writing and balancing equations, we adjust coefficients, the numbers in front of compounds, to ensure that the atoms on both sides of the equation are equal.

For instance, in our textbook exercise, the balanced equation tells us that for every mole of iron(III) oxide that reacts, three moles of carbon monoxide are required to form iron and carbon dioxide. This ratio is key to determining the limiting reactant and ultimately the outcome of the reaction. Mastering chemical reaction balancing requires practice, but it's an essential skill for predicting the amounts of products made and the reactants needed for a given reaction.