A `10g` mixture of `Cu_(2)S` and `CuS` was treated with `200 mL` of `075M MnO_(4)^(-)` in acid solution producing `SO_(2), CU^(2+)` and `Mn^(2+)`. The `SO_(2)` was boiled off and the excess of `MnO_(4)^(-)`. The `SO_(2)` was boiled off and the excess of `MnO_(4)^(-)` was titrated with `175 mL` of `1M Fe^(2+)` solution. Caculaate `%` fo `CuS` in original mixture.

Answer

`57.94%`