5. 2.83 g of a sample of haematite iron ore [iron (II)oxide, Fe2O3] were dissolved in c oncentrated hydrochloric acid and the solution diluted to 250 cm3. 25.0 cm3 of this so lution was reduced with tin(II) chloride (which is oxidised to Sn4+ in the process) to for m a solution of iron(II) ions. This solution of iron(II) ions required 26.4 cm3 of a 0.020 0 mol/dm3 potassium dichromate(VI) solution for complete oxidation back to iron(III) ions. (a) given the half-cell reactions (i) Sn4+ (aq) +2e==> Sn2+ (aq) and (ii) Cr2O72- (aq) + 14H H(aq) + 6e- ==> 2Cr3+(aq) +7H206) deduce the fully balanced redox equations for the reactions (i) the reduction of iron(III) ions by tin(II) ions (ii) the oxidation of iron(II) ions by the dichromate(VI) ion (b) Calculate the percentage of iron (II) oxide in the ore.​

Respuesta :

From the solution of hydrochloric acid that is in the reaction with pottassium dichromate, the fully balanced equation for the reaction is K₂Cr₂O₇+6FeCl + 14HCI→ 2KCl +6FeCl₃ + 2CrCl₃+7H₂0

What is oxidation?

This is the chemical reaction that takes place due to the fact that oxygen has been added to a substance.

a. Fe2O3+ 6HCl→2FeCl₂ + 3H₂O

Fe³⁺+e→Fe2+

2FeCl₃+SnCl₂ → 2FeCl₂ + SnCl₄

Fe²⁺ - e→Fe³⁺

K₂Cr₂O₇+6FeCl + 14HCI→ 2KCl +6FeCl₃ + 2CrCl₃+7H₂0

b. 0.02 mol/dm³ = 2x10⁻⁵ mol/ml

26.4 cm³ is the same as 26.4 ml

n(K₂Cr₂0₇) = 26.4 x2x10⁻⁵

This gives 52.8x10⁻⁵

n(FeCls) = 6n(K₂Cr₂0₇) = 3.168 x 10⁻³

n(FeCl₂)= n(FeCl₃) = 3.168x10⁻³

2 x 56 + 3x16 = 160 gram/mol

3.168x10⁻³/2 =  1.584x10⁻³

1.584x10⁻³ x 160 = 0.253

0.253 x 100/2.83 = 8.96 percent

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