Question

An impure sample of pyrolusite ore consist of 70 of mno2 20 inert impurities and rest moisture. On strong heating , all MnO2 is converted into MnO alongwith formation of O2 . The % of Mn in dried sample is (atomic mass of Mn = 55)

Answer 1

I think Values in Question are like :
70% MnO2 ,
20% Inert Impurities
10% Moisture

Final Answer : 57.37%

STEPS:
1) Let the total mass of Pyrolusite ore be 'x' (in grams.)
Then,
Mass of MnO2 = 0.7 x
Mass of Inert Impurity = 0.2x
Mass of Moisture = 0.1x

2) Reaction :
MnO2 ------> MnO. + 1/2 O2

No. of moles of MnO2
= Given mass / Molar Mass

$= \frac{0.7x}{55 + 2 \times 16} = \frac{0.7x}{87}$
By Reaction,
No. of moles of MnO2 = No. of moles of MnO
= n(MnO)
$\frac{0.7x}{87}$

3) Mass of Mn in MnO = n(MnO) * 55

$= \frac{0.7x}{87} \times 55$

4)
Since, Moisture gets vaporised after heating.
Dried Sample is of : MnO + Inert Impurities
Mass of MnO = n(MnO) * (55+16)

$= \frac{0.7x}{87} \times 71$
5) Finally,
% of Mn in dried Sample :
=
Mass of Mn in MnO /( Mass of MnO + Mass of Inert Impurity)

$= \frac{ \frac{0.7x}{87} \times 55}{ \frac{0.7x}{87} \times 71 + 0.2x} \times 100 \\ \\ = \frac{385}{671} \times 100 \\ \\ = 57 .37 \: \: percentage$

Therefore, % of Mn in dried Sample is
57.37 % .

Answer 2

Answer:

57.4% is ur answer

good luckk