wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A quantity of 4.35 g of a sample of pyrolusite ore, when heated with conc. HCl, gave chlorine. The chlorine when passed through potassium iodide solution, liberated 6.35 g of iodine. The percentage of pure MnO2 in the pyrolusite ore is: (Mn=55,I=127).

A
40
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
50
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
60
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
70
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 50
Given, Pyrolusite means MnO2 ore
The reaction involved is
MnO2+4HClMnCl2+Cl2+2H2O
Again, given that this chlorine is passed through KI
Cl2+2KI2KCl+I2
Here I2 liberated=6.35g
Using the above equations it is possible to find mass of MnO2
No. of moles of I2 liberated=6.35254g/mol=0.025 mol
From abore equation
1 mole MnO21 mol Cl2
& also 1 mole Cl21 mole I2
that means 0.025 mol I2 came from 0.025 mol MnO2 initially present.
In given sample
Mass of MnO2 present=0.025mol×87 g/mol
=2.175g
Purity%=2.175Totalmassofsample×100
=2.175g4.35g×100=50%

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Balancing_oxidation number
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon