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A 0.518 g sample of limestone is dissolv...

A 0.518 g sample of limestone is dissolved in HCl and then the calcium is precipitated as `Ca_(2)C_(2)O_(4)`. After filtering and washing the precipitate, it requires 40 mL of 0.25 N `KMnO_(4)` solution acidified with `H_(2)SO_(4)` to titrate it as,
`MnO_(4)^(-) + H^(+) + C_(2)O_(4)^(2-) to CO_(2) + Mn^(2+) + 2H_(2)O`
The percentage of CaO in the sample is :

A

0.54

B

`27.1%`

C

0.42

D

0.84

Text Solution

Verified by Experts

The correct Answer is:
A
Number of milliequivalents of `CaC_(2)O_(4), KMnO_(4) " and" CaO` will be same.
`40xx0.25=(W)/(56//2)xx1000`
W=0.28 g (Mass of CaO)
`% CaO=(0.28)/(0.518)xx100=54%`
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A 0.518 g sample of limestone is dissolved in HCl and then the calcium is precipitated as CaC_(2)O_(4) . After filtering and washing the precipitate, it requires 40.0 filtering and washing the precipitate, it requires 40.0 mL of 0.250 N KMnO_(4) , solution acidified with H_(2)SO_(4) to titrate it as. The percentage fo CaO in the sample is: MnO_(4)^(-)+H^(+)+C_(2)O_(4)^(2-)rarrMn^(2+)+CO_(2)+2H_(2)O

In the balanced equation MnO_(4)^(-)+H^(+)+C_(2)O_(4)^(2-)rarr Mn^(2+)+CO_(2)+H_(2)O , the moles of CO_(2) formed are :-

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In the reaction C_(2) O_(4)^(-2) + MnO_(4)^(-) + H^(+) rarr Mn^(+2) +CO_(2) the reductants is -

In the reaction C_(2)O_(4)^(2-)+MnO_(4)^(-)+H^(+) rarr Mn^(2+)+CO_(2)+H_(2)O the reductant is

A 0.56 g sample of limestones is dissolved in acid and the calcium is precipitated as calcium oxalate .The precipitate as calcium oxalate the prepcipate is filtered washed with water and dissolved in dil H_(2)SO_(4) The solution required 40ml of 0.25NKmnO_(4) solutions for titration .Calculate percentage of 0.25N KMnO_(4) solution for titration .Calculate of 0.25N KMnO_(4) solution for titration ,Calculate percentage of CaO in limestone sample.

KMnO_(4) react with oxalic acid according to the equation, 2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+) rarr 2Mn^(2+)+ 10 CO_(2)+8H_(2)O , here 20 ml of 0.1 M KMnO_(4) is equivalemt to

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