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.

Inniting MnO_(2) in air converts it quantitctively to Mn_(3)O_(4). A s

1.	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.
Answer» `59.4%` `55%` `56.8%` `58.6%` Solution :`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%`