Which law explain the formation of 5 oxides of nitrogen?
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Answered by
11
The 5 oxides of nitrogen are =
NO, NO₂, N₂O₅, N₂O, N₂O₃
In NO, 1g N combines with 16/14
= 1.14g of oxygen
In NO₂, 1g N combines with 32/14
= 2.28g of oxygen
In N₂O₅, 1g N combines with 80/28
= 2.85g of oxygen
In N₂O, 1g N combines with 16/28
= 0.57g of oxygen
In N₂O₃, 1g N combines with 48/28
= 1.71g of oxygen
So, calculating ratio of masses of oxygen in all above reactions, we get
Ratio = 2 : 4 : 5 : 1 : 3
This is a simple ratio to one another.
Therefore the Law of Multiple Proportions takes place here.
NO, NO₂, N₂O₅, N₂O, N₂O₃
In NO, 1g N combines with 16/14
= 1.14g of oxygen
In NO₂, 1g N combines with 32/14
= 2.28g of oxygen
In N₂O₅, 1g N combines with 80/28
= 2.85g of oxygen
In N₂O, 1g N combines with 16/28
= 0.57g of oxygen
In N₂O₃, 1g N combines with 48/28
= 1.71g of oxygen
So, calculating ratio of masses of oxygen in all above reactions, we get
Ratio = 2 : 4 : 5 : 1 : 3
This is a simple ratio to one another.
Therefore the Law of Multiple Proportions takes place here.
Answered by
3
Hello Dear.
Law of Multiple Proportions explains the Formation of the 5 oxides of the Nitrogen.
____________________________
Explanation ⇒
Five Oxides of the Nitrogen.
NO, NO₂, N₂O₅, N₂O and N₂O₃.
We know,
Mass of N = 14 g.
Mass of Oxygen = 16 g.
In First Case, (NO).
14 g of Nitrogen Combines with 16 g of Oxygen to form NO.
∴ 1 g of Nitrogen Combines with 16/14 g of Oxygen to Form NO.
= 1.14 g of Oxygen.
In Second Case, (NO₂),
Mass of the Oxygen = 16 × 2
= 32 g.
14 g of N is combines with 32 g of Oxygen to form NO₂.
∴ 1 g of N is combines with 32/14 g of Oxygen to form NO₂.
= 2.28 g.
In Third Case,(N₂O₅)
Mass of the Oxygen = 16 × 5
= 80 g.
Mass of Nitrogen = 14 × 2
= 28 g.
∵ 28 g of N is combines with 80 g of Oxygen to form N₂O₅
∴ 1 g of N is combines with 80/28 g of Oxygen to form N₂O₅.
= 2.86 g.
In Fourth Case,(N₂O),
Mass of the Nitrogen = 14 × 2
= 28 g.
∵ 28 g of the N combines with 16 g of the Oxygen to form N₂O.
∴ 1 g of N combines with 16/28 g of the Oxygen to form N₂O.
= 0.57 g.
In Fifth Case, (N₂O₃),
Mass of N = 14 × 2
= 28 g.
Mass of the Oxygen = 16 × 3
= 48 g.
∵ 28 g of N combines to form the 48 g of the Oxygen to form N₂O₃.
∴ 1 g of N combines to form the 48/28 g of the Oxygen to form N₂O₃.
= 1.72 g.
Now, Let us Calculating the Ratio of the masses of the Oxygen.
∴ Ratio = 2 : 4 : 5 : 1 : 3
__________________________
Thus, it is proved that the Law of the Multiple Proportion is followed here.
Hope it helps.
Law of Multiple Proportions explains the Formation of the 5 oxides of the Nitrogen.
____________________________
Explanation ⇒
Five Oxides of the Nitrogen.
NO, NO₂, N₂O₅, N₂O and N₂O₃.
We know,
Mass of N = 14 g.
Mass of Oxygen = 16 g.
In First Case, (NO).
14 g of Nitrogen Combines with 16 g of Oxygen to form NO.
∴ 1 g of Nitrogen Combines with 16/14 g of Oxygen to Form NO.
= 1.14 g of Oxygen.
In Second Case, (NO₂),
Mass of the Oxygen = 16 × 2
= 32 g.
14 g of N is combines with 32 g of Oxygen to form NO₂.
∴ 1 g of N is combines with 32/14 g of Oxygen to form NO₂.
= 2.28 g.
In Third Case,(N₂O₅)
Mass of the Oxygen = 16 × 5
= 80 g.
Mass of Nitrogen = 14 × 2
= 28 g.
∵ 28 g of N is combines with 80 g of Oxygen to form N₂O₅
∴ 1 g of N is combines with 80/28 g of Oxygen to form N₂O₅.
= 2.86 g.
In Fourth Case,(N₂O),
Mass of the Nitrogen = 14 × 2
= 28 g.
∵ 28 g of the N combines with 16 g of the Oxygen to form N₂O.
∴ 1 g of N combines with 16/28 g of the Oxygen to form N₂O.
= 0.57 g.
In Fifth Case, (N₂O₃),
Mass of N = 14 × 2
= 28 g.
Mass of the Oxygen = 16 × 3
= 48 g.
∵ 28 g of N combines to form the 48 g of the Oxygen to form N₂O₃.
∴ 1 g of N combines to form the 48/28 g of the Oxygen to form N₂O₃.
= 1.72 g.
Now, Let us Calculating the Ratio of the masses of the Oxygen.
∴ Ratio = 2 : 4 : 5 : 1 : 3
__________________________
Thus, it is proved that the Law of the Multiple Proportion is followed here.
Hope it helps.
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