Igniting MnO_(2) coverts it quantitatively to Mn_(3)O_(4). A sample of pyrolusite is of the following composition : MnO_(2)=80%,SiO_(2) and other inert contents =15%, rest being water. The sample is ignited in air to constant weight. What is the percentage of Mn in the ignited sample? (Atomic weight of Mn = 55)

Igniting MnO_(2) coverts it quantitatively to Mn_(3)O_(4). A sample of

1.	Igniting MnO_(2) coverts it quantitatively to Mn_(3)O_(4). A sample of pyrolusite is of the following composition : MnO_(2)=80%,SiO_(2) and other inert contents =15%, rest being water. The sample is ignited in air to constant weight. What is the percentage of Mn in the ignited sample? (Atomic weight of Mn = 55)
Answer» Solution :`undersetunderset("=261 g")(3(55+2xx16))(3MnO_(2))rarrundersetunderset("= 229 g")(3xx55+4xx16)(Mn_(3)O_(4))+O_(2)` Suppose the amount of pyrolusite ignited = 100 g `therefore""MnO_(2)" present in it = 80 g"` `SiO_(2)" and other inert contents = 15 g"` `"Water "=100-(80+15)=5g` `"Now,261 g of "MnO_(2)" give "Mn_(3)O_(4)=229g` `therefore"80 g of "MnO_(2)" will give "Mn_(3)O_(4)=(229)/(261)xx80g=70.19g` An on ignition, water is REMOVED but `SiO_(2)` and other inert contents remain, therefore TOTAL mass of the residue after ignition `=70.19+15g=85.19g` `underset(=229g)underset(3xx55+4xx16)(Mn_(3)O_(4))-=underset(=165g)underset(3xx55)(3Mn)` `"229 g "Mn_(3)O_(4)" CONTAIN Mn"=165 g` `therefore""70.19 Mn_(3)O_(4)" contain Mn"=(165)/(229)xx70.19g=50.57g` This is present in 85.19 g of the total residue `therefore""%" of Mn in the residue "=(50.57)/(85.19)xx100=59.36`