Question

A 50.0 g sample of ore containing FeBr₂ is dissolved in acid and titrated to an end point with 24.80 mL of 0.629 M Cr₂O₇²⁻.

The following redox reaction takes place:

Cr₂O₇²⁻ (aq) + Fe²⁺ (aq) ⟶ Cr³⁺ (aq) + Fe³⁺ (aq)

a. balance the reaction

b. what % of the ore sample if FeBr2 (molar mass = 215.65 g/mol)

Answer 1

oxidation half reaction                                             reduction half reaction

Fe^2+(aq) --------> Fe^3+ (aq)                                Cr2O7^2- (aq) -----------> 2Cr^3+ (aq)

Fe^2+(aq) --------> Fe^3+ (aq)                                Cr2O7^2- (aq) -----------> 2Cr^3+ (aq)+7H2O(l)

Fe^2+(aq) --------> Fe^3+ (aq)                                Cr2O7^2- (aq)+14H^+ (aq) -----------> 2Cr^3+ (aq)+7H2O(l)

Fe^2+(aq) --------> Fe^3+ (aq) +e^-                    Cr2O7^2- (aq)+14H^+ (aq)+6e^- ----> 2Cr^3+ (aq)+7H2O(l)

                         6Fe^2+(aq) --------> 6Fe^3+ (aq) +6e^-                

   Cr2O7^2- (aq)+14H^+ (aq)+6e^- ----> 2Cr^3+ (aq)+7H2O(l)

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                Cr2O7^2- (aq) + 6Fe^2+ (aq) +14H^+ (aq) ------------> 6Fe^3+(aq) + 2Cr^3+ (aq) + 7H2O(l)

no of moles of Cr2O7^2-     = molarity * volume in L

                                             = 0.629*0.0248

                                              = 0.0155992moles

from balanced equation

1 mole of Cr2O7^2- react with 6 moles of Fe^2+

0.0155992 moles of Cr2O7^2- react with = 6*0.0155992/1    = 0.0935952 moles of Fe^2+

mass of FeBr2   = no of moles * gram molar mass

                           = 0.0935952*215.65   = 20.184g

%FeBr2 in the ore sample   = 20.184*100/50   = 40.368% >>>>answer