Answer to Question #320859 in General Chemistry for Dickson

Question

2.83 g of a sample of haematite iron ore [iron (III) oxide, Fe2O3] were dissolved in concentrated hydrochloric acid and the solution diluted to 250 cm3. 25.0 cm3 of this solution was reduced with tin(II) chloride (which is oxidised to Sn4+ in the process) to form a solution of iron(II) ions. This solution of iron(II) ions required 26.4 cm3 of a 0.0200 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+(aq) + 6e– ==> 2Cr3+(aq) + 7H2O(l)

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(III) oxide in the ore.

Answer 1

$Fe_2O_3+ 6HCl \rightarrow 2FeCl_3+3H_2O$

$Fe^{3+} +e\rightarrow Fe^{2+}$

$2FeCl_3+SnCl_2 \rightarrow 2FeCl_2+SnCl_4$

$Fe^{2+} -e\rightarrow Fe^{3+}$

$K_2Cr_2O_7 +6FeCl_2 +14HCl \rightarrow 2KCl + 6FeCl_3+2CrCl_3 +7 H_2O$

b. 0.02 mol/dm³=2x10^-5 mole/ml

26.4 cm³=26.4 ml

n(K₂Cr₂O₇)=26.4 x2x10^-5=52.8x10^-5

$n(FeCl_2)=6n(K_2Cr_2O_7)=3.168 x10^{-3}$

$n(FeCl_2)=n(FeCl_3)=3.168x10^{-3}$

$n(Fe_2O_3)=1/2n(FeCl_3)=3.168x10^{-3}/2=1.584 x10^{-3}$

$M(Fe_2O_3)=2x56+3x16=160 \ g/mol$

$m(Fe_2O_3)=1.584x10^{-3} x160 =0.25344 g$

$w(Fe_2O_3)=0.25344x100/2.83=8.96\%$