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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`
A. `54.0%`
B. `27.1%`
C. `42%`
D. `84%`

1 Answer

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Best answer
Correct Answer - A
`M eq. of KMnO_(4)=M eq. of C_(2)O_(4)^(2-)=M eq. of CaCO_(3)`
`40xx.25=` Meq of `CaO`
`(10xx10^(-3))/(2)=` Mole of `CaO`
`%CaO=(5xx10^(-3)xx56xx100)/(518)`
`CaO=54%`

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