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bagirrra123 [75]
2 years ago
13

Calculate the number of milliliters of 0.440 M KOH required to precipitate all of the Fe2+ ions in 187 mL of 0.692 M FeSO4 solut

ion as Fe(OH)2. The equation for the reaction is: FeSO4(aq) + 2KOH(aq) Fe(OH)2(s) + K2SO4(aq)
Chemistry
1 answer:
EleoNora [17]2 years ago
6 0

Answer:

588.2 mL

Explanation:

  • FeSO₄(aq) + 2KOH(aq) → Fe(OH)₂(s) + K₂SO₄(aq)

First we <u>calculate how many Fe⁺² moles reacted</u>, using the given <em>concentration and volume of FeSO₄ solution</em> (the number of FeSO₄ moles is equal to the number of Fe⁺² moles):

  • moles = molarity * volume
  • 187 mL * 0.692 M = 129.404 mmol Fe⁺²

Then we convert Fe⁺² moles to KOH moles, using the stoichiometric ratios:

  • 129.404 mmol Fe⁺² * \frac{2mmolKOH}{1mmolFeSO_4} = 258.808 mmol KOH

Finally we<u> calculate the required volume of KOH solution</u>, using <em>the given concentration and the calculated moles</em>:

  • volume = moles / molarity
  • 258.808 mmol KOH / 0.440 M = 588.2 mL
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What is the molality of a solution if 100.0 g of glucose (C&amp;Hi20e) were dissolved into 750. mL of water?
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