MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 10
        
Igniting MnO2 converts it quantitatively to Mn3O4. A sample of pyrolusite is of the following composition : MnO2 80%, SiO2 and other inert constituents 15%, rest being water. The sample is ignited in air to constant weight. What is the percentage of Mn in the ignited sample?
[O = 16, Mn = 54.9]
5 years ago

Answers : (1)

Jitender Pal
askIITians Faculty
365 Points
							
3 MnO2 → Mn3O4
3 (54.9 + 32) (3 * 54.9 + 64)
= 260.7 g = 228.7 g
Let the amount of pyrolusite ignited = 100.00 g
∴ Wt. of MnO2 = 80 g (80% of 100 g = 80 g)
Wt. of SiO2­ and other inert substances = 15 g
Wt. of water 100 – (80 + 15) = 5 g
According to equation,
260.7 g of MnO2 gives = 228.7 g of Mn3O4
∴ 80 g of MnO2 gives = 228.7/260.7 * 80 = 70.2 g of Mn3O4
NOTE :
During ignition, H2O present in pyrolusite is removed while silica and other inert substances remain as such.
∴ Total wt. of the residue = 70.2 + 15 = 85.2 g
Calculation of % of Mn in ignited Mn3O4
3 Mn = Mn3O4
3 * 54.9 = 164.7 g 3 * 54.9 + 64 = 228.7g
Since, 228.7 g of Mn3O4 contains 164.7 g of Mn
70.2 g of Mn3O4 contains = 164.7/228.7 * 70.2 = 50.55 g of Mn
Weight of resdue = 85.2 g
Hence, percentage of Mn is the ignited sample
= 50.55/85.2 * 100 = 59.33%
5 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 141 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details