
The ammonia \[(N{H_3})\] released on the quantitative reaction of 0.6 g urea \[(N{H_2}CON{H_2})\]with sodium hydroxide \[(NaOH)\] can be neutralised by:
A) \[200ml\,of\,0.2\,N\,HCl\]
B) \[200ml\,of\,0.4\,N\,HCl\]
C) \[100ml\,of\,0.1\,N\,HCl\]
D) \[100ml\,of\,0.2N\,HCl\]
Answer
163.5k+ views
Hint: The quantitative analysis of a reaction helps us to determine the abundance of a substance or substances that are present in a chemical reaction. In a quantitative type of analysis, the amount or concentration of the substance that is analysed is determined and given a numerical value.
Complete Step by Step Solution:
On the reaction of 0.6 grams of urea \[(N{H_2}CON{H_2})\] with sodium hydroxide \[(NaOH)\], ammonia \[(N{H_3})\] is liberated. After balancing the equation, the chemical reaction can be given as :
\[N{H_2}CON{H_2} + 2NaOH \to 2N{H_3} + N{a_2}C{O_3}\]
In the reaction that takes place, one mole of urea gives 2 moles of ammonia when the reaction is balanced. Therefore,
\[{n_{urea}} = \dfrac{{0.6}}{{60}} = 0.01\,\] mole
And, \[{n_{ammonia}} = 2 \times 0.01 = 0.02\] moles
When ammonia is reacted with hydrochloric acid then,
\[N{H_3} + HCl \to N{H_4}Cl\]
0.02 moles of ammonia require a total of 0.02 moles of hydrochloric acid. We know that the n factor of hydrochloric acid is 1, therefore
The molarity of \[HCl\] is equal to 0.2M.
\[0.2 = \dfrac{{{n_{HCl}}}}{{100ml}} \times 1000\]
\[{n_{HCl}} = \dfrac{{0.2 \times 100}}{{1000}} = 0.02\,mol\,of\,HCl\]
So, to neutralise 0.6g of ammonia, 0.02 moles of hydrochloric acid is required.
Hence, option D is the correct answer
Note: The analysis that takes place in a reaction is of two types; Qualitative and Quantitative. Qualitative analysis is based on the physical and chemical properties of a substance in a chemical reaction whereas quantitative analysis helps us to determine the amount or the concentration of a substance in a chemical reaction and is represented by a numerical value. In the qualitative type of analysis, the properties considered are often in the form of odor, color, melting point, boiling point, etc.
Complete Step by Step Solution:
On the reaction of 0.6 grams of urea \[(N{H_2}CON{H_2})\] with sodium hydroxide \[(NaOH)\], ammonia \[(N{H_3})\] is liberated. After balancing the equation, the chemical reaction can be given as :
\[N{H_2}CON{H_2} + 2NaOH \to 2N{H_3} + N{a_2}C{O_3}\]
In the reaction that takes place, one mole of urea gives 2 moles of ammonia when the reaction is balanced. Therefore,
\[{n_{urea}} = \dfrac{{0.6}}{{60}} = 0.01\,\] mole
And, \[{n_{ammonia}} = 2 \times 0.01 = 0.02\] moles
When ammonia is reacted with hydrochloric acid then,
\[N{H_3} + HCl \to N{H_4}Cl\]
0.02 moles of ammonia require a total of 0.02 moles of hydrochloric acid. We know that the n factor of hydrochloric acid is 1, therefore
The molarity of \[HCl\] is equal to 0.2M.
\[0.2 = \dfrac{{{n_{HCl}}}}{{100ml}} \times 1000\]
\[{n_{HCl}} = \dfrac{{0.2 \times 100}}{{1000}} = 0.02\,mol\,of\,HCl\]
So, to neutralise 0.6g of ammonia, 0.02 moles of hydrochloric acid is required.
Hence, option D is the correct answer
Note: The analysis that takes place in a reaction is of two types; Qualitative and Quantitative. Qualitative analysis is based on the physical and chemical properties of a substance in a chemical reaction whereas quantitative analysis helps us to determine the amount or the concentration of a substance in a chemical reaction and is represented by a numerical value. In the qualitative type of analysis, the properties considered are often in the form of odor, color, melting point, boiling point, etc.
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