Problem 97 - General Chemistry: Principles and Modern Applications

Question

Hydrogen gas, \(\mathrm{H}_{2}(\mathrm{g}),\) is passed over \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})\) at \(400^{\circ} \mathrm{C} .\) Water vapor is formed together with a black residue-a compound consisting of \(72.3 \% \mathrm{Fe}\) and \(27.7 \%\) O. Write a balanced equation for this reaction.

Step-by-Step Solution

Answer

The balanced equation for the reaction is \(3H_{2}(g) + Fe_{2}O_{3}(s) \rightarrow 2FeO(s) + 3H_{2}O(g)\)

Step 1: Convert Percent Composition to Mole Ratios of the Unknown Compound

To convert percentages by weight in a compound to mole ratios, it is necessary first to convert the weight percentages to grams. Assuming a 100g sample, then there will be 72.3g of Fe and 27.7g of O. Since mole ratios are required, these weights are then converted to moles using the molar mass of each element. The molar mass of Fe is approximately \(55.85 g/mol\) while for O, it's \(16.00 g/mol\). Thus, the number of moles of Fe is \(72.3g/55.85 g/mol \approx 1.29 mol\) and for O it's \(27.7g/16.00g/mol \approx 1.73 mol\).

Step 2: Determine the Empirical Formula of the Unknown Compound

Empirical formula represents the ratio of atoms in a compound, which is derived from the mole amounts of its constituent elements. Dividing each of these mole quantities by the smallest of them will give the empirical formula. Thus, as \(1.29 mol/1.29 mol \approx 1\) and \(1.73 mol/1.29 mol \approx 1.34\), we can't have fractional amounts in an empirical formula therefore we round the numbers to the closest whole number. Thus, the empirical formula for the compound is FeO.

Step 3: Determine the Stoichiometric Coefficients and Write the Balanced Equation

Hydrogen reacts with iron(III) oxide to produce the compound and water. The reaction is written as follows: \(H_{2}(g) + Fe_{2}O_{3}(s) \rightarrow FeO(s) + H_{2}O(g)\). To balance this equation it is evident that there should be 3 molecules of \(H_2\) and 1 molecule of \(Fe_{2}O_{3}\) on the left hand side as one molecule of \(Fe_{2}O_{3}\) can give 2 molecules of 'Fe' necessary for 2 molecules of \(FeO\), and 3 molecules of \(H_2\) can give 3 molecules of \(H_2O\). However, 2 molecules of \(FeO\) gives 2 'O' atoms which along with the one 'O' atom from \(H_{2}O\) gives a total of 3 'O' atoms matching the left hand side. This means the balanced reaction would be: \(3H_{2}(g) + Fe_{2}O_{3}(s) \rightarrow 2FeO(s) + 3H_{2}O(g)\)

Key Concepts

Empirical Formula

Understanding the empirical formula is key in determining the simplest ratio of elements in a compound. In our problem, we are given a residue resulting from a reaction, composed of iron (Fe) and oxygen (O). The challenge here is to find the empirical formula based on the percentage composition given: 72.3% iron and 27.7% oxygen.

To start, we assume a 100-gram sample, making our percentages equal to grams. This assumption simplifies calculations. We convert these masses into moles by using each element's molar mass:

• For iron, with a molar mass of 55.85 g/mol, we have 72.3 g / 55.85 g/mol ≈ 1.29 moles of Fe.


• For oxygen, with a molar mass of 16.00 g/mol, we calculate 27.7 g / 16.00 g/mol ≈ 1.73 moles of O.

Next, we determine the simplest whole number ratio of moles by dividing each by the smallest number of moles, which in this case is 1.29. This yields 1 for Fe and 1.34 for O. To translate these into whole numbers for the empirical formula, approximate 1.34 to the closest whole number, which is 1, resulting in the empirical formula FeO for the compound found in the residue.

Stoichiometry

Stoichiometry is essential to quantify relative amounts of reactants and products in a chemical reaction. In the exercise, once the empirical formula (FeO) is identified, we apply stoichiometry to balance the entire chemical equation of the reaction.

From the problem, hydrogen (\(H_2(g)\)) reacts with iron(III) oxide (\(Fe_2O_3(s)\)) to form iron oxide (FeO) and water (\(H_2O(g)\)). An unbalanced equation for the reaction would start as:\[H_{2}(g) + Fe_{2}O_{3}(s) \rightarrow FeO(s) + H_{2}O(g)\]

Stoichiometry helps to determine coefficients that balance this equation, ensuring that the number of each type of atom is the same on both sides, obeying the law of conservation of mass. Let's break down the parts:

• Iron: \(Fe_2O_3\) must yield two Fe atoms for two FeO.


• Oxygen: \(Fe_2O_3\) has three O atoms, while products side has \(2 \times FeO + H_2O\) needing three O atoms.


• Hydrogen: To balance, consider hydrogens in \(H_2\) and \(H_2O\), needing three \(H_2\) to match three O atoms forming three \(H_2O\).

In conclusion, stoichiometry provides a framework to decipher the exact quantities and relationship among reactants and products, crucially confirming the final balanced equation.

Balancing Chemical Equations

Balancing chemical equations ensures that the same quantity of atoms for each element is observed on both sides of the reaction. According to the exercise, our goal is to balance the chemical equation for the reaction between hydrogen gas and iron(III) oxide.

A balanced chemical equation properly reflects the law of conservation of mass. Let’s consider our unbalanced equation:\[H_{2}(g) + Fe_{2}O_{3}(s) \rightarrow FeO(s) + H_{2}O(g)\]

Balance the equation by adjusting coefficients:

• Realize two Fe from \(Fe_2O_3\) yield two FeO. Thus, use a coefficient of 2 for FeO.


• For oxygen, three oxygen in \(Fe_2O_3\) need distribution across two \(FeO\) and another O for three \(H_2O\) formation, dictating a coefficient of 3 for \(H_2O\).


• Ensuring hydrogens match three \(H_2O\) compounds, results in a coefficient of 3 for \(H_2(g)\):

The final balanced equation becomes:\[3H_{2}(g) + Fe_{2}O_{3}(s) \rightarrow 2FeO(s) + 3H_{2}O(g)\]

To balance equations effectively, always start by balancing elements that appear in only one reactant and one product. Adjust coefficients as needed, ensuring every element is balanced. It’s often helpful to leave hydrogen and oxygen last, as they are frequently part of multiple compounds. This balancing act beautifies science, demonstrating perfect symmetry and conservation!