`2.36 g` of sample of dolomite containing only `CaCO_(3)` and `MgCO_(3)` were dissolved in `700 mL` of `0.1 N HCl`. The solution was dilutied to `2.05 L`. `25 mL` of this solution required `20 mL` of `0.01 N NaOH` solution for complete neutralisation. Find the percent composition of ore.

Explanation

`[12.36 g (CaCO_(3)` and `MgCO_(3)) + 700 mL of 0.1 NHCl]` in a `2.5 L` solution
`implies` mEq of `HCl` used for dolomite `= 70 - 20 = 50`
`25 mL` of above (diluted) sample `-= 20 mL` of `0.01 N NaOH` Note that the sample contains `CaCO_(3)` and `MgCO_(3)`, which are bases, so both will be neutralised. (Both of these carbonates react as diacidic bases)
The diluted sample contains excess of `HCl` solution.
`implies` mEq of excess `HCl = mEq of NaOH`
`("in" 25 mL)` `0.01 xx 20 = 0.2`
`implies` mEq of excess `HCl` in `2.5 L = 0.2 xx (2500)/(25) = 20`
mEq of `HCl` (added to dolomite) `= 0.1 xx 700 = 70`
Let `x g` of `CaCO_(3)` and `(2.36 - x) g` of `MgCO_(3)` be present in dolomite.
`implies (mEq of CaCO_(3) + mEq of MgCO_(3)) = 50`
`implies` solve for `x`.
`x = 1.625 g =` mass of `CaCO_(3)`
and `(2.36 - x) = 0.73 g =` mass of `MgCO_(3)`
`implies` Percentage of `CaCO_(3) = 68.8%`
and percentage of `MgCO_(3) = 31.2%`