Math Problem Statement

Identify the oxidizing and reducing agents in the given reactions and calculate the molarity of a diluted solution.

Solution

Solution for the Questions:


b) Identify the oxidizing and reducing agents:

  1. Reaction (i): I2O5+5CO5CO2+I2I_2O_5 + 5CO \rightarrow 5CO_2 + I_2

    • Oxidation: Carbon in COCO increases oxidation state from +2+2 to +4+4 in CO2CO_2. Therefore, COCO is the reducing agent.
    • Reduction: Iodine in I2O5I_2O_5 decreases oxidation state from +5+5 to 00 in I2I_2. Therefore, I2O5I_2O_5 is the oxidizing agent.
  2. Reaction (ii): 2Fe2++H2O2+2H+2Fe3++2H2O2Fe^{2+} + H_2O_2 + 2H^+ \rightarrow 2Fe^{3+} + 2H_2O

    • Oxidation: Iron Fe2+Fe^{2+} increases oxidation state from +2+2 to +3+3. Therefore, Fe2+Fe^{2+} is the reducing agent.
    • Reduction: Oxygen in H2O2H_2O_2 decreases oxidation state from 1-1 to 2-2 in H2OH_2O. Therefore, H2O2H_2O_2 is the oxidizing agent.

c) Molarity of the resulting solution:

  1. Step 1: Molar mass of NaFNaF:

    • Sodium (Na): 22.99 g/mol
    • Fluorine (F): 18.998 g/mol
    • Total: 22.99+18.998=41.988g/mol22.99 + 18.998 = 41.988 \, \text{g/mol}
  2. Step 2: Calculate moles of NaFNaF: Moles of NaF=Mass of NaFMolar mass of NaF=125.6g41.988g/mol=2.99mol\text{Moles of NaF} = \frac{\text{Mass of NaF}}{\text{Molar mass of NaF}} = \frac{125.6 \, \text{g}}{41.988 \, \text{g/mol}} = 2.99 \, \text{mol}

  3. Step 3: Molarity of the original solution:

    • Volume = 1 L Molarity=Moles of soluteVolume of solution=2.99mol1L=2.99M\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution}} = \frac{2.99 \, \text{mol}}{1 \, \text{L}} = 2.99 \, \text{M}
  4. Step 4: Calculate moles in 180.0 mL of the solution: Moles=Molarity×Volume (in L)=2.99M×0.180L=0.5382mol\text{Moles} = \text{Molarity} \times \text{Volume (in L)} = 2.99 \, \text{M} \times 0.180 \, \text{L} = 0.5382 \, \text{mol}

  5. Step 5: Molarity of the diluted solution:

    • Total volume after dilution = 500 mL = 0.500 L Molarity of diluted solution=Moles of soluteVolume of solution=0.5382mol0.500L=1.076M\text{Molarity of diluted solution} = \frac{\text{Moles of solute}}{\text{Volume of solution}} = \frac{0.5382 \, \text{mol}}{0.500 \, \text{L}} = 1.076 \, \text{M}

Final Answers:

  1. b (i):

    • Oxidizing agent: I2O5I_2O_5
    • Reducing agent: COCO
  2. b (ii):

    • Oxidizing agent: H2O2H_2O_2
    • Reducing agent: Fe2+Fe^{2+}
  3. c:

    • Molarity of the resulting solution: 1.076M1.076 \, \text{M}

Do you want further explanations or clarifications?

Related Questions:

  1. What is the difference between oxidizing and reducing agents in chemical reactions?
  2. How do you identify the oxidation states of elements in a compound?
  3. What is the significance of molarity in chemical reactions?
  4. How would the molarity change if the solution were further diluted?
  5. What practical applications involve identifying oxidizing and reducing agents?

Tip: Always confirm oxidation states by calculating for each element systematically to avoid errors.

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Math Problem Analysis

Mathematical Concepts

Oxidation-Reduction Reactions
Solution Chemistry
Stoichiometry

Formulas

Molar mass = Sum of atomic masses of elements in a compound
Moles = Mass / Molar Mass
Molarity = Moles of Solute / Volume of Solution
Dilution Formula: M1V1 = M2V2

Theorems

Law of Conservation of Mass

Suitable Grade Level

Grades 11-12