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Chapter 4: Problem 81

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)? \(4 \mathrm{Fe}(\mathrm{OH})^{+}(a q)+4 \mathrm{OH}^{-}(a q)+\mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$4 \mathrm{Fe}(\mathrm{OH})_{3}(s)$$

Short Answer

Expert verified
Answer: Approximately 0.054 grams of O鈧� are consumed.

Step by step solution

01

Calculate moles of iron ions present

To calculate moles of iron ions in the 75 mL solution, use the equation: moles = molarity 脳 volume where, molarity = 0.090 M volume = 75 mL = 0.075 L moles of Fe(OH)鈦� = 0.090 M 脳 0.075 L = 0.00675 mol
02

Use stoichiometry to find moles of O鈧� needed

According to the balanced chemical equation: 4 Fe(OH)鈦� + 4 OH鈦� + O鈧� + 2 H鈧侽 鈫� 4 Fe(OH)鈧� 4 moles of Fe(OH)鈦� react with 1 mole of O鈧�. Thus, the moles of O鈧� needed can be calculated as follows: moles of O鈧� = (moles of Fe(OH)鈦� 脳 1) / 4 moles of O鈧� = (0.00675 mol 脳 1) / 4 = 0.0016875 mol
03

Convert moles of O鈧� to grams

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 脳 volume in liters
Plugging in the given values, we get:
  • moles of Fe(OH)鈦� = 0.090 M 脳 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)鈦� + 4 OH鈦� + O鈧� + 2 H鈧侽 鈫� 4 Fe(OH)鈧�
This balanced equation conveys that 4 moles of Fe(OH)鈦� react with 1 mole of O鈧�. 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 O鈧�, which is 32 g/mol since each oxygen atom has a molar mass of 16 g/mol and O鈧� consists of two oxygen atoms.

To convert moles of O鈧� to grams, we use the formula:
  • mass = moles 脳 molar mass
In our example, with 0.0016875 moles of O鈧� required, the conversion is:
  • mass of O鈧� = 0.0016875 mol 脳 32 g/mol 鈮� 0.054 grams
This step provides the final answer to how much O鈧� 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.

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Most popular questions from this chapter

When electrodes connected to a lightbulb are inserted into a beaker containing silver carbonate and water, will the bulb not glow, glow dimly, or glow brightly? What do you think will happen after addition of one equivalent of aqueous HCl? Write a balanced net ionic equation that supports your answer.

What is the difference between a strong base and a weak base?

Give the oxidation number of nitrogen in each of the following: (a) elemental nitrogen \(\left(\mathrm{N}_{2}\right) ;\) (b) hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right) ;(\mathrm{c})\) ammonium ion \(\left(\mathrm{NH}_{4}^{+}\right)\)

Which of the following reactions of calcium compounds is \(/\) are redox reactions? a. \(\quad \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\) b. \(\mathrm{CaO}(s)+\mathrm{SO}_{2}(g) \rightarrow \mathrm{CaSO}_{3}(s)\) c. \(\mathrm{CaCl}_{2}(s) \rightarrow \mathrm{Ca}(s)+\mathrm{Cl}_{2}(g)\) d. \(3 \mathrm{Ca}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Ca}_{3} \mathrm{N}_{2}(s)\)

When a solution of dithionate ions \(\left(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}\right)\) is added to a solution of chromate ions \(\left(\mathrm{CrO}_{4}^{2-}\right),\) the products of the reaction under basic conditions include soluble sulfite ions and solid chromium(III) hydroxide. This reaction is used to remove chromium(VI) from wastewater generated by factories that make chrome-plated metals. a. Write the net ionic equation for this redox reaction. b. Which element is oxidized and which is reduced? c. Identify the oxidizing and reducing agents in this reaction. d. How many grams of sodium dithionate would be needed to remove the chromium(VI) in \(100.0 \mathrm{L}\) of wastewater that contains \(0.00148 M\) chromate ion?
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