Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

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
`        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

Jitender Pal
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 ?

## Other Related Questions on Physical Chemistry

View all Questions »

### 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