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zysi [14]
3 years ago
12

Iron(II) can be oxidized by an acidic K2Cr2O7 solution according to the net ionic equation: Cr2O72− + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe

3+ + 7H2O If it takes 35.5 mL of 0.0250 M K2Cr2O7 to titrate 25.0 mL of a solution containing Fe2+, what is the molar concentration of Fe2+ in the original solution?
Chemistry
1 answer:
cestrela7 [59]3 years ago
4 0

Answer:

The molar concentration of Fe²⁺ in the original solution is 1.33 molar

Explanation:

Moles of K₂Cr₂O₇ = Molarity x Volume (lit)

                             = 0.025 x 35.5 x 10⁻³

                             = 0.0008875

Cr₂O₇²⁻ + 6 Fe²⁺ + 14 H⁺ → 2 Cr³⁺ + 6 Fe³⁺ + 7 H₂O

From equation

              1 mole K₂Cr₂O₇ used for the oxidation of 6 moles Fe²

0.0008875 mole K₂Cr₂O₇ used for the oxidation of =\frac{6 X 0.0008875}{1} = 0.005325 mole of  Fe²

      Molarity = \frac{No. of moles of solute}{Volume     of solution(lit)}

Molar concentration of  Fe² = \frac{0.005325 X 1000}{25} = 1.33 molar

So molar concentration of Fe²⁺ in the original solution = 1.33 molar

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<h3>What are the required values of sides and angles?</h3>

Here in the question it is given that,

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Hence DA = 13, BW = 5, WC = 12, ∠ACD = 25°, ∠DAB = 50°, ∠ADC = 130°, ∠DBC = 65° and ∠BWC = 90° given that ABCD is a rhombus and DB = 10, BC = 13, ∠WAD = 25°.

Learn more about rhombus here:

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