
White fumes are:
(A) Chlorine
(B) Hydrogen sulphide
(C) Ammonium chloride
(D) Ammonium hydroxide
Answer
530.4k+ views
Hint: Knowing the compound at the place of ‘A’ can help us solve the given equations easily and thus, find the answer.
As, we need to find only the composition of white fumes, we can solve only one side of the given equation. But for better understanding we should simplify each side of the reaction given (as finding ‘B’ would be simpler from the other path as $A\to C\to B$ )
Complete answer:
Let us move towards the given sample of equation to simplify it and find the required answer.
By analysing and having basic knowledge, let us say that the compound ‘A’ is ammonium nitrate i.e. $N{{H}_{4}}N{{O}_{3}}$ . Now, by moving step by step;
1. On thermal decomposition i.e. heating ammonium nitrate, it gives a mixture of ${{N}_{2}}O+{{H}_{2}}O$ as,
$N{{H}_{4}}N{{O}_{3}}\to {{N}_{2}}O+2{{H}_{2}}O$
2. On heating ammonium nitrate with caustic soda, it gives a mixture of $NaN{{O}_{3}}+N{{H}_{3}}+{{H}_{2}}O$ as,
$N{{H}_{4}}N{{O}_{3}}+NaOH\to NaN{{O}_{3}}+N{{H}_{3}}+{{H}_{2}}O$
Then the mixture of $NaN{{O}_{3}}+N{{H}_{3}}+{{H}_{2}}O$ on reduction with Zn powder gives a gas which is a replacement of ‘B’ i.e. ammonia gas.
3. On heating ammonium nitrate with excess NaOH, it gives ‘B’ i.e. ammonia gas along with other products. When the ammonia gas produced reacts with HCl, it forms white fumes of ammonium chloride as,
$N{{H}_{3}}+HCl\to N{{H}_{4}}Cl$
Thus, we can say that,
A – ammonium nitrate
B – ammonia gas
White fumes produced – ammonium chloride
All this can be shown by a sample equation as,
Therefore, option (C) is correct.
Note:
Do note to analyse each and every step to have a brief understanding of the whole equation and thus the reaction taking place.
As, we need to find only the composition of white fumes, we can solve only one side of the given equation. But for better understanding we should simplify each side of the reaction given (as finding ‘B’ would be simpler from the other path as $A\to C\to B$ )
Complete answer:
Let us move towards the given sample of equation to simplify it and find the required answer.
By analysing and having basic knowledge, let us say that the compound ‘A’ is ammonium nitrate i.e. $N{{H}_{4}}N{{O}_{3}}$ . Now, by moving step by step;
1. On thermal decomposition i.e. heating ammonium nitrate, it gives a mixture of ${{N}_{2}}O+{{H}_{2}}O$ as,
$N{{H}_{4}}N{{O}_{3}}\to {{N}_{2}}O+2{{H}_{2}}O$
2. On heating ammonium nitrate with caustic soda, it gives a mixture of $NaN{{O}_{3}}+N{{H}_{3}}+{{H}_{2}}O$ as,
$N{{H}_{4}}N{{O}_{3}}+NaOH\to NaN{{O}_{3}}+N{{H}_{3}}+{{H}_{2}}O$
Then the mixture of $NaN{{O}_{3}}+N{{H}_{3}}+{{H}_{2}}O$ on reduction with Zn powder gives a gas which is a replacement of ‘B’ i.e. ammonia gas.
3. On heating ammonium nitrate with excess NaOH, it gives ‘B’ i.e. ammonia gas along with other products. When the ammonia gas produced reacts with HCl, it forms white fumes of ammonium chloride as,
$N{{H}_{3}}+HCl\to N{{H}_{4}}Cl$
Thus, we can say that,
A – ammonium nitrate
B – ammonia gas
White fumes produced – ammonium chloride
All this can be shown by a sample equation as,
Therefore, option (C) is correct.
Note:
Do note to analyse each and every step to have a brief understanding of the whole equation and thus the reaction taking place.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

