Animated Solution for Chemistry - Basic Concepts in Chemistry: The ammonia (${\text{NH}}_3$) released on quantitative reaction of $0.6\text{ g}$ urea (${\text{NH}}_2{\text{CONH}}_2$) with sodium hydroxide ($\text{NaOH}$) can be neutralised by

A Stoichiometric Journey

Imagine you are in a laboratory, and you are tasked with neutralizing a basic gas produced from a chemical reaction. This is exactly what this problem asks us to do. We are given a specific amount of urea, and we need to find out how much hydrochloric acid (HCl) is required to neutralize the ammonia gas it produces when reacted with sodium hydroxide.

The Chemical Reaction

The very first step in any stoichiometry problem is to write down the balanced chemical equation. When urea (${\text{NH}}_2{\text{CONH}}_2$) reacts quantitatively with sodium hydroxide ($\text{NaOH}$), it undergoes a hydrolysis-like reaction in an alkaline medium to release ammonia gas (${\text{NH}}_3$) and form sodium carbonate (${\text{Na}}_2{\text{CO}}_3$).

The balanced equation is:

Notice the stoichiometry: One mole of urea yields exactly two moles of ammonia. This $1:2$ ratio is the critical bridge that connects the reactant to the product.

Calculating Moles

Before we can talk about the ammonia produced, we need to know exactly how much urea we started with in terms of moles. We are given the mass of urea as $0.6\text{ g}$.

The molar mass of urea can be calculated by adding the atomic masses of its constituent atoms: $\text{Molar mass}=14(\text{N})+2(\text{H})+12(\text{C})+16(\text{O})+14(\text{N})+2(\text{H})=60\text{ g/mol}$.

Now, let's find the moles of urea:

Moles of Ammonia

Using the stoichiometric ratio we identified earlier, we can now determine the moles of ammonia produced. Since $1\text{ mole}$ of urea gives $2\text{ moles}$ of ammonia, $0.01\text{ moles}$ of urea will give:

This $0.02\text{ mol}$ of ammonia is the base that we need to neutralize using our acid, HCl.

The Law of Equivalence

To completely neutralize a base with an acid, the Law of Equivalence states that the number of equivalents of the acid must be equal to the number of equivalents of the base.

For ammonia, the n-factor (valency factor) is $1$ because it can accept one proton (${\text{H}}^{+}$) to form the ammonium ion (${\text{NH}}_4^{+}$). Therefore, its equivalents are equal to its moles:

This means we need exactly $0.02\text{ equivalents}$ of HCl. The equivalents of a solution can be calculated using the formula:

Let's test the given options to find the perfect match: - Option (a): $100\text{ mL}$ of $0.2\text{ N HCl}⟹0.2\times (100\times 10^{-3})=0.02\text{ eq}$. This is a perfect match! - Option (b): $200\text{ mL}$ of $0.4\text{ N HCl}⟹0.4\times (200\times 10^{-3})=0.08\text{ eq}$. (Too much acid) - Option (c): $200\text{ mL}$ of $0.2\text{ N HCl}⟹0.2\times (200\times 10^{-3})=0.04\text{ eq}$. (Too much acid) - Option (d): $100\text{ mL}$ of $0.1\text{ N HCl}⟹0.1\times (100\times 10^{-3})=0.01\text{ eq}$. (Not enough acid)

Thus, the exact amount of acid required is $100\text{ mL}$ of $0.2\text{ N HCl}$.