Question

A.) FeCl2 is water soluble. what ions are formed when this compound dissolves in water?

B.) If the solution is 0.255 M FeCl2 what is the molarity of each ion in aqueous solution?

C.) Find the percentage of FeCl2 in a 35.5g sample of ore which has been dissolved in aqueous acid and titrated to the endpoint with 23.55 ml of 0.309 M Cr2O7^2- . the molar mass of FeCl2 is 126.0 g/mol. Cr2O7^2- + Fe^2+ ------> Cr^3+ + Fe^3+ (acidic solution)

D.) how many mL of 0.122M MnO4^-1 are required to react with an excess of C2O4^2- in order to produce 75.25 mL of carbon dioxide gas measured at 30.0 degree c and 715 torr?

MnO4^-(aq) + C2O4^2-(aq) ......> MnO2(s) +CO2(g) (basic solution )

Answer 1

A) FeCl2(s) + H2O -------> Fe²⁺(aq) + 2Cl^-(aq)

B) Molarity of Fe²⁺ = 0.255M

Molarity of Cl^- = 2* 0.255M = 0.510M

C) Balanced equation is

Cr2O7^2- +6 Fe³⁺ -------> 2Cr³⁺ + 6Fe²⁺

1mole of Cr2O7^2- react with 6mol of Fe²⁺

No of mole of Cr2O7^2- consumed = (0.309 mol / 1000ml0* 23.55ml = 0.007277

No of mole of Fe²⁺ = 0.007277*6=0.043662

molar mass of Fe²⁺ = 126.0g/mol

Mass of mole of FeCl2 = 126.0g/mol * 0.043662mol = 5.5014g

% of FeCl2 in sample = (5.5014g/35.5g)*100= 15.50%

D) No of mole of CO2,n = PV/RT

= 0.9438atm * 0.07525L/0.082057( l atm / mol K ) 303.15K

= 0.002855mol

Partly balanced equation is

2MnO4^- + 3C2O4^2- --------> MnO2(s) + 6CO2(g)

6 mole of CO2 is produced from 2mole of MnO4^-

therefore

No of mole of MnO4^- reacted =( 2/6)* 0.002855 = 0.0009517

Volume containing 0.0009517 mole of MnO4^- = (1000/0.122mole)*0.0009517 = 7.8ml