Question

0.56g of limestone was treated with oxalic acid to give CaC2O4. The precipitate decolourised 45ml of 0.2N KMnO4 in acid medium. The % of CaO in limestone is:

Answer 1

                                    oxalic acid                        KMnO₄
Limestone [ CaCO₃ ] -----------------→ CaC₂O₄ --------------→ decolorizes

Redox changes :

For CaC₂O₄          (C ³⁺)₂                 →           2C⁴⁺        +             2e

For KMnO4            5e           +             Mn⁷⁺       →           Mn²⁺

Meq. of CaCO₃ = Meq. of CaO = Meq. of KMnO₄
[As, Meq. of CaCO₃ = Meq. of CaO, because CaO is present in CaCO₃]

Meq. of CaO = Meq. of KMnO₄

So, w ÷ (56/2) × 1000 = 45 × 0.2          [w = weight of CaO]

w = 0.252g

Therefore, percentage of CaO = (0.252 / 0.56) × 100

Percentage = 45%

Answer 2

Answer:

We have,  

Lime stone          oxalic acid                   KMnO4                  

CaCO3                   →                       CaC2O4             →                      decolourizes

Therefore,          redox changes are :

For CaC2O4                          (C3+)2                 →           2C4+        +             2e  

For KMnO4         5e           +             Mn7+       →           Mn2+

Meq. of CaCO3 =             Meq. of CaO      =             Meq. of KMnO4  

[As, Meq. of CaCO3 = Meq. of CaO, because CaO is present in CaCO3]

Therefore,          Meq. of CaO      =             Meq. of KMnO4

Or,                      w/(56/2) × 100 =             45 × 0.2                 [w = weight of CaO]

Or,                       w            =             0.252 g  

Therefore,          percentage of CaO          =             (0.252/0.56) × 100%

=             45%                 (C)[Ans.]

Explanation: