Problem 81 - Chemistry The Science in Context

Question

Iron(II) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, which converts soluble \(\mathrm{Fe}(\mathrm{OH})^{+}\) to insoluble \(\mathrm{Fe}(\mathrm{OH})_{3} .\) How many grams of \(\mathrm{O}_{2}\) are consumed to precipitate all of the iron in \(75 \mathrm{mL}\) of \(0.090 M\) iron(II)? \(4 \mathrm{Fe}(\mathrm{OH})^{+}(a q)+4 \mathrm{OH}^{-}(a q)+\mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$4 \mathrm{Fe}(\mathrm{OH})_{3}(s)$$

Step-by-Step Solution

Answer

Answer: Approximately 0.054 grams of O₂ are consumed.

Step 1: Calculate moles of iron ions present

To calculate moles of iron ions in the 75 mL solution, use the equation: moles = molarity × volume where, molarity = 0.090 M volume = 75 mL = 0.075 L moles of Fe(OH)⁺ = 0.090 M × 0.075 L = 0.00675 mol

Step 2: Use stoichiometry to find moles of O₂ needed

According to the balanced chemical equation: 4 Fe(OH)⁺ + 4 OH⁻ + O₂ + 2 H₂O → 4 Fe(OH)₃ 4 moles of Fe(OH)⁺ react with 1 mole of O₂. Thus, the moles of O₂ needed can be calculated as follows: moles of O₂ = (moles of Fe(OH)⁺ × 1) / 4 moles of O₂ = (0.00675 mol × 1) / 4 = 0.0016875 mol

Step 3: Convert moles of O₂ to grams

To convert moles of O₂ to grams, use the equation: mass = moles × molar mass where, moles of O₂ = 0.0016875 mol molar mass of O₂ = 32 g/mol (16 g/mol × 2, since there are 2 oxygen atoms in O₂) mass of O₂ = 0.0016875 mol × 32 g/mol ≈ 0.054 g So, approximately 0.054 grams of O₂ are consumed to precipitate all of the iron in 75 mL of 0.090 M iron(II) solution.

Key Concepts

Molarity Calculations

Molarity is a way to express the concentration of a solution. It tells us how many moles of a solute are present in one liter of solution. This measure is essential for understanding how much of a substance is in the solution, which is crucial when conducting chemical reactions. In this topic, we calculate molarity by multiplying the given molarity by the volume of the solution in liters.

For example, if we have a solution with a molarity of 0.090 M and a volume of 75 mL, we first need to convert the volume from milliliters to liters. This conversion is done by dividing the milliliters by 1000, resulting in 0.075 L.

The number of moles of iron ions present can be calculated using the formula:

• moles = molarity × volume in liters

Plugging in the given values, we get:

• moles of Fe(OH)⁺ = 0.090 M × 0.075 L = 0.00675 moles

These calculations are critical for the following steps in stoichiometric problems.

Balanced Chemical Equations

Balanced chemical equations are fundamental in chemistry. They illustrate how reactants transform into products, ensuring that the number of atoms of each element is conserved. This means the equation reflects the law of conservation of mass. Each side of the equation mirrors the other in terms of the number of atoms per element.

In our particular reaction, we have:

• 4 Fe(OH)⁺ + 4 OH⁻ + O₂ + 2 H₂O → 4 Fe(OH)₃

This balanced equation conveys that 4 moles of Fe(OH)⁺ react with 1 mole of O₂. Balancing equations is critical to finding out how much of each reactant is required: it forms the basis of stoichiometry, allowing us to calculate the moles needed and predict the amounts produced in reactions.

The concept of stoichiometry emerges from this, helping us derive how much of a product can form from given reactants, creating a bridge between known quantities and unknowns in chemical reactions.

Moles to Grams Conversion

Converting moles to grams is a common step in stoichiometry, as we often need to know the mass of a substance involved in a reaction. The conversion uses the molar mass of a compound, which is the mass of one mole of that substance, typically found on the periodic table.

The step in this context involves knowing the molar mass of O₂, which is 32 g/mol since each oxygen atom has a molar mass of 16 g/mol and O₂ consists of two oxygen atoms.

To convert moles of O₂ to grams, we use the formula:

• mass = moles × molar mass

In our example, with 0.0016875 moles of O₂ required, the conversion is:

• mass of O₂ = 0.0016875 mol × 32 g/mol ≈ 0.054 grams

This step provides the final answer to how much O₂ is necessary, making it an essential part of solving chemical reaction equations. Understanding this process helps in accurately preparing and measuring substances in laboratory setups.