Click to Chat

1800-2000-838

+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
Menu
pallavi pradeep bhardwaj Grade: 12
`        A metal oxide is reduced by heating it in a stream of hydrogen.It is found that after compelete reduction 3.15g of the oxide have yielded 1.05g of the metal.We may conclude that`
8 years ago

## Answers : (2)

Ramesh V
70 Points
```										consider +1 and +2 oxidation states for metals
case 1 :
M2O + H2  -- > 2M  +H2O
2x+16                 2x
so, its , 2*3.15/(2x+16) = 1.05/x
or x= 4 i.e., the compound may be Helium(but helium being inert gas doesn't for oxide)
case 2 :
MO + H2  -- > M  +H2O
x+16                 x
so, its , 3.15/(x+16) = 1.05/x
or x = 8 (but there is no element with atomic weight of 8)
--
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it
here and we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE
preparation. All the best.

Regards,
Ramesh

```
8 years ago
sakshi
11 Points
```										actually it can be much more easier . we all know that there is no metal with atomic mass of 8 and helium doesn’t form oxide . so the answer can be calculated by the following method:mass of metal/massof oxygen(both given in ques.)= eq. mass of metal (E) /eq.mass of oxygen (i.e.8)1.05 / (3.15-1.05) =E/8E=4
```
6 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

## Other Related Questions on Physical Chemistry

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details

## Ask Experts

Have any Question? Ask Experts

Post Question

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