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%`

A `0.518 g` sample of limestone is dissolved in `HCl` and then the cal

1.	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%`
Answer» Correct Answer - A 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%`

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