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)
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
A 50.0 g sample of ore containing FeBr2 is dissolved in acid and titrated to an...
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