Problem 42 - Chemistry: Principles and Reactions

Question

Rust, which you can take to be \(\mathrm{Fe}(\mathrm{OH})_{3}\), can be dissolved by treating it with oxalic acid. An acid-base reaction occurs, and a complex ion is formed. (a) Write a balanced equation for the reaction. (b) What volume of \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) would be required to remove a rust stain weighing \(1.0 \mathrm{~g}\) ?

Step-by-Step Solution

Answer

1. Write the balanced equation: 2Fe(OH)3 + 6H2C2O4 → 2Fe(C2O4)3³⁻ + 6H2O. 2. Calculate the moles of rust (1.0 g) using its molar mass: 0.00935 mol Fe(OH)3. 3. Determine the moles of oxalic acid needed based on the balanced equation: 0.02805 mol H2C2O4. 4. Calculate the volume of 0.10 M oxalic acid solution required to provide this amount: 280.5 mL of 0.10 M H2C2O4.

Step 1: (a) Writing the balanced equation

First, we need to write the chemical reaction between rust (\(\mathrm{Fe}(\mathrm{OH})_{3}\)) and oxalic acid (\(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\)). In this acid-base reaction, the rust will react with the acid, and the complex ion \(\mathrm{Fe}(\mathrm{C}_{2}\mathrm{O}_{4})_{3}^{3-}\) will be formed along with water. The unbalanced equation can be written as: $$ \mathrm{Fe}(\mathrm{OH})_{3} + \mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} \rightarrow \mathrm{Fe}(\mathrm{C}_{2}\mathrm{O}_{4})_{3}^{3-} + \mathrm{H}_{2}\mathrm{O} $$ To balance the equation, we can add coefficients in front of the reactants and products: $$ 2\mathrm{Fe}(\mathrm{OH})_{3} + 6\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} \rightarrow 2\mathrm{Fe}(\mathrm{C}_{2}\mathrm{O}_{4})_{3}^{3-} + 6\mathrm{H}_{2}\mathrm{O} $$ Thus, the balanced equation for the reaction is: $$ 2\mathrm{Fe}(\mathrm{OH})_{3} + 6\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} \rightarrow 2\mathrm{Fe}(\mathrm{C}_{2}\mathrm{O}_{4})_{3}^{3-} + 6\mathrm{H}_{2}\mathrm{O} $$

Step 2: (b) Calculating the volume of oxalic acid solution

To find the volume of \(0.10 \mathrm{M} \mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\) solution required to remove a 1.0 g rust stain, follow these steps: 1. Calculate the moles of rust (\(\mathrm{Fe}(\mathrm{OH})_{3}\)) in 1.0 g: Molar mass of \(\mathrm{Fe}(\mathrm{OH})_{3} = 1\times(55.85) + 3\times(16.00+1.01) = 106.87 \mathrm{~g/mol}\). Moles of rust: \(\frac{1.0\mathrm{~g}}{106.87\mathrm{~g/mol}} = 0.00935 \mathrm{~mol}\). 2. Calculate the moles of oxalic acid (\(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\)) needed to react with this amount of rust by using the balanced equation: For every 2 moles of rust, we need 6 moles of oxalic acid: Moles of oxalic acid: \(0.00935 \mathrm{~mol} \times \frac{6}{2} = 0.02805 \mathrm{~mol}\). 3. Determine the volume of \(0.10 \mathrm{M} \mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\) required to provide this number of moles: Volume: \(\frac{0.02805 \mathrm{~mol}}{0.10 \mathrm{~M}} = 0.2805 \mathrm{~L}\) or \(280.5 \mathrm{~mL}\). So, to remove 1.0 g of rust stain, we need \(280.5 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\) solution.

Key Concepts

Balanced Chemical Equation

In chemical reactions, the balanced chemical equation is the essential tool that represents the conservation of mass, meaning the number of atoms for each element in the reactants equals those in the products. In our example, we are examining the reaction between rust, represented by \( \mathrm{Fe} ( \mathrm{OH} )_3 \), and oxalic acid, which is \( \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \). The first step is to write the unbalanced chemical equation:\[ \mathrm{Fe} ( \mathrm{OH} )_3 + \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightarrow \mathrm{Fe}(\mathrm{C}_2 \mathrm{O}_4)_3^{3-} + \mathrm{H}_2 \mathrm{O} \] Here, iron hydroxide reacts with oxalic acid to form an iron oxalate complex ion and water.To balance this equation, adjust the coefficients:- Place a '2' in front of \( \mathrm{Fe} ( \mathrm{OH} )_3 \) and \( \mathrm{Fe}(\mathrm{C}_2 \mathrm{O}_4)_3^{3-} \), since both contain iron.- Place a '6' in front of \( \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \) and \( \mathrm{H}_2 \mathrm{O} \) because they need to match the number of oxalate ions and hydrogen atoms.The balanced equation is:\[ 2\mathrm{Fe} ( \mathrm{OH} )_3 + 6\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightarrow 2\mathrm{Fe}(\mathrm{C}_2 \mathrm{O}_4)_3^{3-} + 6\mathrm{H}_2 \mathrm{O} \]

Molar Mass Calculation

Knowing how to calculate molar mass is key to converting between grams and moles. Here's how we calculate the molar mass of a compound, like rust, or \( \mathrm{Fe} ( \mathrm{OH} )_3 \). It involves summing up the atomic masses of all atoms present:- \( \mathrm{Fe} \) accounts for 55.85 g/mol.- \( \mathrm{O} \) contributes 16.00 g/mol each, thus, for three oxygen atoms, it's 48.00 g/mol.- \( \mathrm{H} \) adds 1.01 g/mol per hydrogen atom, resulting in an additional 3.03 g/mol for three hydrogens.Add these components:\[ 55.85 + (3 \times 16.00) + (3 \times 1.01) = 106.87 \text{ g/mol} \]To find moles of \( \mathrm{Fe} ( \mathrm{OH} )_3 \) in 1.0 g: Use the formula:\[ \text{Moles} = \frac{1.0 \text{ g}}{106.87 \text{ g/mol}} \approx 0.00935 \text{ mol} \] This step is crucial for quantitative chemical analysis and reactions.

Complex Ion Formation

Complex ion formation describes the creation of a larger structure from simpler ions, which in this case is the formation of \( \mathrm{Fe}(\mathrm{C}_2 \mathrm{O}_4)_3^{3-} \) when iron combines with oxalate ions.These structures are a result of coordinate covalent bonds, where ligands (like oxalate) donate electron pairs to a metal center, which is the iron.- Each rust molecule, \( \mathrm{Fe} ( \mathrm{OH} )_3 \), releases iron which interacts with oxalic acid through a series of reactions.- As the acid breaks down in solution, oxalate ions (\( \mathrm{C}_2 \mathrm{O}_4^{2-} \)) are liberated.- Iron centers, represented here as \( \mathrm{Fe}^{3+} \), bind with these oxalate ions, resulting in the complex ion \( \mathrm{Fe}(\mathrm{C}_2 \mathrm{O}_4)_3^{3-} \).Understanding this concept helps visualize why certain reactions result in larger and more complex products, and it's fundamental in areas like coordination chemistry used in many industrial processes.