All the carbon atom present in KH3(C2O4)2.2H2O weighting 254 gm is converted to CO2. How many gram of CO2 were obtained.

 Illustrative solution:

Related Questions:

Principle of atom conservation (PAOC):

Illustrative example (2) 27.6 gm K2CO3 (138) was treated by a series of reagents so as to convert all its carbon into K2Zn3[Fe(CN)6]2(MM=698) Calculate mass of K2Zn3[Fe(CN)6]2.

Illustrative example (3)0.32 gm mole of LiAlH4 in ether solution was placed in a flask and 74 gm (1 mole ) of t- butyl alcohol was added . The product is LiAlHC12H27O3. Calculate mass of product if all lithium atoms are converted.

0.32 gm mole of LiAlH4 in ether solution was placed in a flask and 74 gm (1 mole ) of t- butyl alcohol was added . The product is LiAlHC12H27O3. Calculate mass of product if all lithium atoms are converted.

Illustrative solution:

By applying POAC on (C) atoms:

Related Questions:

Principle of atom conservation (PAOC):

Illustrative example (2) 27.6 gm K2CO3 (138) was treated by a series of reagents so as to convert all its carbon into K2Zn3[Fe(CN)6]2(MM=698) Calculate mass of K2Zn3[Fe(CN)6]2.

Illustrative example (3) All the carbon atom present in KH3(C2O4)2.2H2O weighting 254 gm is converted to CO2. How many gram of CO2 were obtained.

Illustrative example (4) For BaCl2.XH2O , 2.1 gm of Compound gives 2 gm of anhydrous BaSO4 (M.mass =233)upon treatment with H2SO4 then calculate value of X.

Illustrative example (5) In a determination of P an aqueous solution of NaH2PO4 is treated with a mixture of Ammonium and magnesium ions to precipitate magnesium Ammonium phosphate (Mg(NH4)PO4.6H2O. This is heated and decomposed to magnesium pyrophosphate. Which is weighed. A solution of NaH2PO4 yield 3.33 gm of Mg2P2O7. What is weighed of NaH2PO4 was present originally.

27.6 gm K2CO3 (138) was treated by a series of reagents so as to convert all its carbon into K2Zn3[Fe(CN)6]2(MM=698) Calculate mass of K2Zn3[Fe(CN)6]2.

 Here all the carbon of K2CO3 converted into a complex hence POAC can apply

Related Questions:

Illustrative example (2)0.32 gm mole of LiAlH4 in ether solution was placed in a flask and 74 gm (1 mole ) of t- butyl alcohol was added . The product is LiAlHC12H27O3. Calculate mass of product if all lithium atoms are converted.

Illustrative example (3) All the carbon atom present in KH3(C2O4)2.2H2O weighting 254 gm is converted to CO2. How many gram of CO2 were obtained.

Illustrative example (4) For BaCl2.XH2O , 2.1 gm of Compound gives 2 gm of anhydrous BaSO4 (M.mass =233)upon treatment with H2SO4 then calculate value of X.

Illustrative example (5) In a determination of P an aqueous solution of NaH2PO4 is treated with a mixture of Ammonium and magnesium ions to precipitate magnesium Ammonium phosphate (Mg(NH4)PO4.6H2O. This is heated and decomposed to magnesium pyrophosphate. Which is weighed. A solution of NaH2PO4 yield 3.33 gm of Mg2P2O7. What is weighed of NaH2PO4 was present originally.

The standard reduction potential for Cu2+/Cu is +0.3.4 volt calculate the reduction potential at PH =14 for above couple. ksp fo Cu(OH)2 is 1.0x10^-19.

Metal-Metal- Insoluble salts electrode:

These consist of a metal (M) which is covered by a layer of sparingly soluble salt (MX), immersed in solution containing common ions (X-). And it is represented as anode or anion electrode.

M I MX (sat’d) I X^-  for example  Ag I AgCl (sat’d) I Cl^-  II  Cathode

We Know that AgX (F, Cl, Br,I) Silver salt’s are insoluble in water.

Other examples:

(1) Hg(s) I Hg₂SO₄(sat’d) I SO₄^-2 II    (Cathode)  used in the standard Weston Cell.

(2) Hg(s) I Hg₂Cl₂(sat’d) I Cl- (KCl paste) II    (Cathode) used as Calomel electrode

(3) Pb(s) I PbSO4(sat’d) I SO₄^-2 II    (Cathode)  used in lead batteries

(4) Cu(s) I CuCl₂(sat’d) I Cl- (KCl paste) II    (Cathode)

(1) Silver electrode as Anode or anionic electrode:

Ag I AgCl (sat’d) I Cl^-  II  Cathode

The electrode reaction can be considered as taking place in two steps as.

This electrode reaction involve only the concentration of chloride ion, the electrode is said to be reversible with respect to chloride ions (Cl-).

The above reaction show how an electron can be released or taken up in such electrode.

Nernst equation for this electrode (Anode) is written as:

Silver electrode as cathode or cationic electrode:

   Anode   II Cl^- aq I AgCl(sat’d)IAg(s)

Nernst equation for this electrode (Cathode) is written as:

(2) Calomel electrode:

A calomel cell consist of a platinum electrode dipping into mercury in contact with calomel (dimercury (I) chloride,Hg₂Cl₂ and potassium chloride solution.

As Anode:

Hg(s) I Hg₂Cl₂(sat’d) I Cl- (KCl paste) II    (Cathode) :

As Cathode:

  Anode          II Cl- (KCl paste)I Hg₂Cl₂(sat’d) I Hg(s)    

Related Questions:

(1) The conductivity of 0.0011028 mol per liter acetic acid is 4.95 × 10–5 S per cm. Calculate its dissociation constant if ^°m for acetic acid is 390.5 S cm2 per mol–1 .

(2) In the electrolysis of aqueous solution of NaOH , 2.8 litre of oxygen at NTP liberated at anode. How much Hydrogen was liberated at cathode ?

(3) On passing 3 faradays of electricity through three electrolytic cells connect in series containing Ag+ , Ca+2 and Al+3 ions respectively , the molar ratio in which three metals ions are liberated at the electrode is

(4) How to determine anode and cathode of concentration cell instead of both electrodes are same ?

(5) Write correct Nernst equation for the given galvanic cell containing both the electrodes are M/M (insoluble) salt at 25°c.: Ag(s)|AgCl(s)| KCl(M) || KBr (M)|AgBr(s)|Ag(s): [Given Ksp(AgCl) and Ksp(AgCl)]

(6) A silver electrode is immersed in saturated solution of Ag2SO4(aq). The potential difference between the silver and standard electrode is found to be 0.711 V. determined Ksp(Ag2SO4) given E0Ag+/Ag= 0.799 V

(7) How many grams of silver could be plated out on a serving tray by electrolysis of solution containing silver in +1 oxidation state for a period of 8 hours. At a current of 8.46 ampere? What is the area of tray if the thickness of silver plating is 0.00254 cm. density of silver is 10.5 gm cc.

(8) Calculate the emf of the cell: Pt H2(1atm)ICH3COOH(0.1M) II NH4OH(0.01M)IH2(1 atm)Pt and Ka for CH3COOH= 1.8x10^-5 and Kb for NH4OH = 1.8x10^-5

(9) A current of 3 ampere passed for 2 hours through a solution of CuSO4 3 gm of Cu2+ ions were discharged at cathode calculate the current efficiency.

(10) Two weak acid solution HA1 and HA2 each with same concentration and having Pka values 3 and 5 are placed in contact with hydrogen electrode ( 1 atm at 25 degree) and are interconnected through a salt bridge find the emf.