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?

Options

A. `50.96 %`
B. `60.48 %`
C. `59.36 %`
D. `65.63 %`

Answer

C

Explanation

`3 MnO_(2) rarr Mn_(3)O_(4) + O_(2)`
From stoichiometry, we have:
3 mol of `MnO_(2) -= 1 mo of Mn_(3)O_(4) -= 2 mol of Mn`
`implies 1 mol of MnO_(2) -= 1//3 mol of Mn_(3)O_(4)`
Let `100 g` of pyrolusite sample be taken.
`implies` grams of `MnO_(2) = 80`
`implies` or moles of `MnO_(2) = (80)/(87)`
From stoichiometry: Moles of `Mn_(3) O_(4) = (1)/(3) ((80)/(87))`
Moles of `Mn = (80)/(87)`
Grams of `Mn = (80)/(87) xx 85 = 50.57 g`
Grams of `Mn_(3) O_(4) = (1)/(3) ((80)/(87)) xx 259 = 70.19`
Total weight of ignited sample
`= 70.19 + 15 (SiO_(2)` and inert matter) `= 85.19`
`implies % Mn = (50.57)/(85.19) xx 100 = 59.36 %`