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`

Options

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

Answer

A

Explanation

No of milli equivalents of lime stone =
no of milli equivalents of `CaC_(2)O_(4) =`
no of milli equivalents of `KMnO_(4) = 40 xx 0.25 = 10`
No of milli moles of lime stone =
No of milli moles of lime =5
`:.` weight of `CaO = 5 xx 56 xx 10^(-3) = 0.28g`
percentage of `CaO = (0.28)/(0.518) xx 100 = 54.05%`