Inniting `MnO_(2)` in air converts it quantitctively to `Mn_(3)O_(4)`. A sample of pyrolusite is of the following composition. `MnO_(2)=80%` and othe inert constituents =15% and rest bearing `H_(2)O`. The sample is ignited to constant weight. What is the % of Mn is the igrited sample.

Options

A. `59.4%`
B. `55%`
C. `56.8%`
D. `58.6%`

Answer

A

Explanation

`3MnO_(2)rarrMn_(3)O_(4)+O_(2)`
`therefore` 1 mol of `MnO_(2)=1//3` mol of `Mn_(3)O_(4)`
Let 100 gm of pyrolusite sample be taken.
Then grams of `MnO_(2)=80` gm
implies moles of `MnO_(2)=80//87`
`therefore` Moles of `Mn_(3)O_(4)=(1)/(3)((80)/(87))`
Moles of `Mn=(80)/(87)=1` wt of `Mn=(80)/(87)xx55`
= 50.57 gm
wt of `Mn_(3)O_(4)=(1)/(3)((80)/(87))xx229=70.19`
Total weight of ignited sample
`=70.19+15=85.19`
`%Mn=(50.57)/(85.19)+100=59.36%`