Igniting `MnO_(2)` in air converts it quantitatively to `Mn_(3)O_(4)`. A sample of pyrolusite is of the following composition: `MnO_(2) = 80%`, `SiO_(2)` and other inert constituents = 15%, and rest bearing `H_(2) O`. The sample is ignited to constant weight. What is the percent of `Mn` in the ignited sample?

Explanation

`underset(261 g)(3MnO_(2)) to underset(229 kg)(Mn_(3)O_(4)) + O_(2)`
Let the amount of pyrolusite ignited = 100 g
Wt. of `MnO_(2) = 80 g`
Wt. of `SiO_(2)` and other inert contents = 15 g
Wr. Of water = 100 - (80 + 15) = 5 g
Now, 261 g of `MnO_(2)` give `Mn_(3)O_(4) = 229`g
80 g of `MnO_(2)` gives `Mn_(2)O_(4) = 229/261 xx 80`
=70.19 g
During ignition, water present in pyrolusite is removed while `SiO_(2)` and other inert contents remain as such.
Total wt. of residue = 70.19 + 15 = 85.19 g
`therefore` Percentage of `Mn_(3)O_(4)` is the residue `=(70.19)/(85.19) xx 100 = 82.39%`
Now, `underset(3 xx 55 = 165)(3 Mn) = underset(3 xx 55 + 4 xx 16=229)(Mn_(3)O_(4))`
`therefore 229` g of `Mn_(3)O_(4)` contain Mn = 165
82.39 g of `Mn_(3)O_(4)` contain Mn `=165/229 xx 82.39`
= 59.36 g
`therefore` Purecentage of Mn in ignited sample = 59.36%