Question

Looking at the stoichiometry of the reaction (see the lab manual), how many moles KMnO₄ must have been delivered by the buret to react with that much Fe(NH₄)₂(SO₄)₂ · 6H₂O?
8H⁺ + 5Fe⁺² + MnO₄^- --> Mn⁺² + 5Fe⁺³ + 4H₂O
(Remember that all Group I salts are soluble - including KMnO₄. Therefore moles KMnO₄ = moles MnO₄^-)

	1.59×10−03 moles
	5.92×10−04 moles
	2.96×10−03 moles
	5.03×10−04 moles

Considering the moles and volume of KMnO₄ above, what must the true molarity of your KMnO₄ solution be?

	0.020 M
	2.26×10−02 M
	7.54×10−03 M
	1.49×10−02 M

Answer 1

8H⁺ + 5Fe⁺² + MnO4 ^- $\rightarrow$ Mn⁺² + 5Fe⁺³ + 4H₂O

Mass of Fe(NH4)2(SO4)2 · 6H2O = 1.01 g

Molar mass of Fe(NH4)2(SO4)2 · 6H2O = 392.16 g/mol

Moles of Fe(NH4)2(SO4)2 · 6H2O = 1.01 / 392.16

= 0.0025

Moles of Fe⁺² = 0.0025

From the equation:

5 moles Fe⁺² reacts with 1 mole MnO₄^-

0.0025 moles Fe⁺² will reacts with = 0.0025 / 5

= 0.0005 moles MnO₄^-

Therefore, moles of KMnO4 needed = 5.03x10^-4 moles.

Volume of KMnO4 used = 25.68 - 0.15

= 25.53 mL = 0.025 L

Molarity of KMnO4 = 5.03x10^-4 / 0.025

= 0.020 M