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

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 )

Explanation / Answer

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

Cr2O72- +6 Fe3+ -------> 2Cr3+ + 6Fe2+

1mole of Cr2O72- react with 6mol of Fe2+

No of mole of Cr2O72- 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- + 3C2O42- --------> 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

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