Question

An oxide of iodine (I=127 atoms) contains 42.3g of iodine and 8g of oxygen. Its formula could be. Solution
plz answer it fast...

Answer 1

Answer:

Correct option is

C

I

2

O

5

Given, 25.4 g iodine combines with 8 g oxygen.

Therefore, 254 g iodine binds with 80 g oxygen.

The ratio of moles I:O=2:5

So, oxide is I

2

O

5

.

Explanation:

GAMER

Answer 2

Given:

Mass of iodine in the oxide of iodine $=42.3 g$

Mass of oxygen in the oxide of iodine $=8g$

To Find: Formula of oxide of iodine

Solution:

Number of moles can be given as:

$\[No.\,of\,moles = \frac{{Mass}}{{Molar\,mass}}\]$

Therefore,

Number of moles of iodine in the oxide of iodine

$\[No.\,of\,moles = \frac{{42.3\,g}}{{127}}\]$

$\[No.\,of\,moles = \,0.3\]$

Number of moles of oxygen in the oxide of iodine

$\[No.\,of\,moles = \frac{{8\,g}}{{16}}\]$

$\[No.\,of\,moles = \,0.5\]$

The ratio of the number of moles of iodine and oxygen can be given as:

$\[\frac{I}{O} = \frac{{0.3}}{{0.5}}\]$

$\[ \Rightarrow I:O = 3:5\]$

Therefore, the formula of the oxide

$\[{I_x}{O_y} = {I_3}{O_5}\]$

Hence, the formula of the oxide of iodine is $\[{I_3}{O_5}\]$.