Calcualte the mass of `95%` pure `MnO_(2)` to produce 35.5 g of `Cl_(2)` as per the following reaction. `MnO_(2) + 4HCl to MnCl_(2) + Cl_(2) + H_(2)O`. (At. Mass of Mn = 55).

Explanation

`MnO_(2) + 4HCl to MnCl_(2) + Cl_(2) + H_(2)O`
`=(55+32) g = 87 g`
71 g of `Cl_(2)` is produced by 87 g of `MnO_(2)`.
35.5 g of `Cl_(2)` is produced by `87/71 xx 35.5= 43.49` g of `MnO_(2)`.
Since, `MnO_(2)` is `95%` pure i.e., 95 g of `MnO_(2)` is present in 100 g of `MnO_(2)` sample.
43.49 g of pure `MnO_(2)` will be present in `=(43.49 xx 100)/95= 45.8 g` of `MnO_(2)` sample.