Problem 71 - Chemistry: An Atoms-Focused Approach

Question

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

Step-by-Step Solution

Answer

Answer: 0.054 grams

Step 1: Determine Moles of Fe²⁺ in the solution

The given solution has a volume of 75 mL and a molarity of 0.090 M. We can convert the volume to liters and use the molarity to calculate the moles of Fe²⁺. $$ Volume (L) = \frac{75mL}{1000} = 0.075 L $$ $$ Moles \, of \, Fe^{2+} = Molarity \times Volume = 0.090 M \times 0.075 L = 0.00675\, moles $$

Step 2: Calculate Moles of O₂ Required

According to the balanced chemical equation provided, the stoichiometry tells us that 4 moles of Fe(OH)²⁺ react with 1 mole of O₂. We can use this ratio to determine the moles of O₂ required to precipitate all the Iron. $$ Moles \, of \, O_{2} = \frac{0.00675 \,moles \, of \, Fe^{2+}}{4} = 0.0016875 \, moles \, of \, O_{2} $$

Step 3: Convert Moles of O₂ into Grams

To convert moles of O₂ to grams, we use the molar mass of O₂, which is 32 g/mol. $$ Grams \, of \, O_{2} = 0.0016875\, moles \times 32 \frac{g}{mol} = 0.054 \, g $$ So, 0.054 grams of \(\mathrm{O}_{2}\) are consumed to precipitate all of the iron in 75 mL of 0.090 M \(\mathrm{Fe}^{2+}\).

Key Concepts

Stoichiometry

Understanding stoichiometry is essential for working with chemical reactions, such as the one in our exercise. Stoichiometry is the part of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In simpler terms, it helps us figure out how much of each substance is needed or produced in a reaction.

For example, the balanced chemical equation in our exercise shows how four moles of \( \mathrm{Fe(OH)}^+ \) and oxygen are used to produce four moles of \( \mathrm{Fe(OH)}_3 \). This balanced equation provides the mole ratio necessary to solve the problem.

• Step one involves using stoichiometry to determine how many moles of each reactant are involved.
• We used this ratio to know that 4 moles of \( \mathrm{Fe}^{2+} \) require 1 mole of \( \mathrm{O}_2 \) for the reaction to proceed.

This basic stoichiometry allows chemists to predict how much of a substance is needed to react completely, or how much product will be formed if the reaction goes to completion.

Molar Calculations

Molar calculations are critical in chemistry for converting between moles, mass, and number of particles. In our exercise, molar calculations facilitated the transition from a macroscopic scale (grams and liters) to a molecular one (moles).

First, we calculated the number of moles of \( \mathrm{Fe}^{2+} \) present in the solution using the formula:

\[\text{Moles} = \text{Molarity} \times \text{Volume} \]

This equation allowed us to find that there are 0.00675 moles of \( \mathrm{Fe}^{2+} \) in 75 mL of a 0.090 M solution. Next, using the stoichiometry from our balanced equation, we determined the amount of \( \mathrm{O}_2 \) in moles needed for the reaction.

Finally, we used the molar mass of \( \mathrm{O}_2 \) (32 g/mol) to convert the moles into grams, which is a more practical unit for measuring on a lab scale. This conversion is essential to answer the question of how much \( \mathrm{O}_2 \) is consumed during the process.

Chemical Equations

Chemical equations illustrate chemical reactions using symbols and formulas to represent substances involved. These equations provide crucial insights and directions for carrying out chemical reactions.

In our example, the chemical equation:
\[4 \mathrm{Fe(OH)}^{+}(a q) + 4 \mathrm{OH}^{-}(a q) + \mathrm{O}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 4 \mathrm{Fe(OH)}_{3}(s)\]
helps us understand which reactants are required and what products are formed.

• This equation shows that the oxygen gas \( \mathrm{O}_2 \) reacts with \( \mathrm{Fe}^{2+} \) ions in a basic solution.
• The equation must be balanced to obey the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.

The balancing of chemical equations ensures the input and output masses are equal, which is fundamental for accurate stoichiometric and molar calculations. By understanding chemical equations, students can efficiently solve problems like determining the quantities of reactants needed or products formed in a given chemical context.