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 ingnited sample?

Explanation

Suppose the weight of pyrolusite is x g
wt. Of `MnO_(2)""( 80)/( 100) xx x = 0.8 x `
Wt. of `SiO_(2) ` etc. = `(15)/(100) xx x = 0.15 x`
Wt. of water ` = (5)/( 100) xx x = 0.05 x`
When pyrolusite is ignited, `MnO_(2)` changes to `Mn_(3)O_(4) and H_(2)O` evaporates.
The residue contains, therefore , `SiO_(2), ` etc., and `Mn_(2) O_(4)`
Now , we know
`MnO_(2) to Mn_(2) O_(4)`
0.8x g
Applying POAC for Mn atoms,
moles of Mn is `MnO_(2)` = moles of Mn in `Mn_(3) O_(4)`
`1 xx` moles of `MnO_(2) = 3 xx ` moles of `Mn_(3) O_(4)" " . . . (i)`
`(0.8)/( 87) = 3 xx ("wt.of "Mn_(3) O_(4))/( 229) [{:(MnO_(2)=87),(Mn_(3)O_(4)=229):}] `
Wt. of `Mn_(3)O_(4) = 0.702` x g
`:.` wt. of the residue = wt. of `Mn_(3)O_(4)` + wt. of `SiO_(2)`, etc.
`= 0.702 x + 0.15 x = 0.852 x g `
Now, since Mn atoms are conserved,
moles of Mn = moles of Mn = `Mn_(3) O_(4)`
= moles of Mn in `MnO_(2)`
`= 1 xx ` moles of `MnO_(2)`
`= (0.8x)/( 87) (MnO_(2) = 87)`
`:. ` wt. of Mn = moles of Mn ` xx ` at . wt. of Mn
`= (0.8 x)/( 87) xx 55 g`
`%` of Mn in residue = `("wt. of Mn")/( "wt. of residue ") xx 100`
`= (0.8 x 55)/(87) xx (100)/( 0.852 x) = 59.37% `