Question

In an oxidation-reduction reaction, 0.0450 mol of aqueous FeSO4 (source of Fe2+) reacts completely with 120.0 mL of an acidified aqueous solution of KMnO4 (source of MnO4-). 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(ℓ) What is the molarity of the KMnO4 solution?

Answer 1

Answer:

0.075 M

Explanation:

5Fe²⁺(aq) + MnO₄⁻(aq) + 8H⁺(aq) → 5Fe³⁺(aq) + Mn²⁺(aq) + 4H₂O(ℓ)

Using the moles of Fe²⁺ that reacted, we can calculate the reacting moles of MnO₄⁻:

0.0450 mol Fe⁺² * $\frac{1molMnO_{4}^{-}}{5molFe^{+2}}$ = 0.0090 mol MnO₄⁻

Now we divide the moles of MnO₄⁻ by the volume in order to calculate the molarity of the solution, keeping in mind that 120.0 mL = 0.120 L.

0.0090 mol MnO₄⁻ / 0.120 L = 0.075 M

Answer 2

The molarity of the KMnO₄ solution : 0.075 M

Further explanation

Stochiometry in Chemistry learns about chemical reactions mainly emphasizing quantitative, such as calculations of volume, mass, number, which are related to the number of actions, molecules, elements, etc.

In chemical calculations, the reaction can be determined, the number of substances that can be expressed in units of mass, volume, mole, or determine a chemical formula, for example, the substance level or molecular formula of the hydrate.

The reaction equation is the chemical formula of reagents and product substances

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products

From the reaction:

5Fe²⁺ (aq) + MnO₄⁻ (aq) + 8H⁺ (aq) → 5Fe³⁺ (aq) + Mn²⁺ (aq) + 4H₂O (ℓ)

From the reaction equation above, the reaction coefficient shows the mole ratio of reagents and products

mole ratio for Fe²⁺: MnO₄⁻ = 5: 1

mole Fe²⁺ = 0.045

then mole MnO₄⁻:

= 0.045 : 5

= 0.009

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

$\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}$

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

then :

$\displaystyle M=\frac{0.009~mole}{0.12~L}\\\\M=0.075$

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Keywords: mole, Fe²⁺, MnO₄⁻, molarity