Animated Solution for Chemistry - s and p-Block Elements: Find $A$, $B$ and $C$ in the following reactions: $N{H}_3+A+C{O}_2⟶(N{H}_4{)}_{2}C{O}_3$ $(N{H}_4{)}_{2}C{O}_3+H_{2}O+B⟶N{H}_{4}HC{O}_3$ $N{H}_{4}HC{O}_3+NaCl⟶N{H}_{4}Cl+C$

The world of industrial chemistry is filled with elegant sequences of reactions that transform simple, abundant raw materials into highly valuable products. The problem we are tackling today is a classic example of such a sequence. It walks us through the foundational steps of the Solvay Process, a brilliant chemical pathway designed to manufacture sodium carbonate (washing soda) on a massive scale.

Instead of just memorizing the reactions, let's dive deep into the physical and chemical realities of what is happening in the reaction vessels. Imagine you are an engineer overseeing these massive tanks. Let's break down the chemistry step by step.

What are we looking at?

The Solvay process relies on a few cheap and easily accessible raw materials: brine (a concentrated solution of sodium chloride, $NaCl$), limestone (calcium carbonate, $CaC{O}_3$), and ammonia ($N{H}_3$). The ultimate goal is to produce sodium carbonate ($N{a}_{2}C{O}_3$), but the journey there involves several intermediate steps. The reactions provided in the question represent the core chemical transformations of this process.

The Birth of Ammonium Carbonate

Let's look at the first reaction in our sequence:

Here, ammonia gas ($N{H}_3$) and carbon dioxide gas ($C{O}_2$) are being brought together to form ammonium carbonate, $(N{H}_4{)}_{2}C{O}_3$. But wait, if you just mix dry ammonia and dry carbon dioxide gases, you won't get this salt. The formation of the carbonate ion ($C{O}_3^{2-}$) requires an aqueous medium.

When ammonia dissolves in water, it forms ammonium hydroxide ($N{H}_{4}OH$), which provides the necessary $N{H}_4^{+}$ and $O{H}^{-}$ ions. Simultaneously, carbon dioxide dissolves in water to form carbonic acid ($H_{2}C{O}_3$), which provides $H^{+}$ and $C{O}_3^{2-}$ ions. These ions then combine to form ammonium carbonate.

Therefore, the missing component A must be water ($H_{2}O$). The complete balanced equation is:

The Shift to Bicarbonate

Now, let's move to the second reaction:

In this step, the ammonium carbonate solution is further treated to form ammonium bicarbonate ($N{H}_{4}HC{O}_3$). Notice the chemical change: we are moving from a carbonate ($C{O}_3^{2-}$) to a bicarbonate ($HC{O}_3^{-}$). How do we achieve this?

By adding more carbon dioxide! When excess $C{O}_2$ is bubbled through the aqueous solution of ammonium carbonate, it reacts with the water to form more carbonic acid. This additional acid provides protons ($H^{+}$) that attach to the carbonate ions, converting them into bicarbonate ions.

This means the missing component B is carbon dioxide ($C{O}_2$). The complete balanced equation is:

The Crucial Ion Exchange

Finally, we arrive at the third reaction, which is the heart of the Solvay process:

Here, the ammonium bicarbonate solution is mixed with a concentrated solution of sodium chloride ($NaCl$), commonly known as brine. What happens next is a classic double displacement reaction. The ions swap partners: the ammonium ion ($N{H}_4^{+}$) pairs up with the chloride ion ($C{l}^{-}$) to form ammonium chloride ($N{H}_{4}Cl$), and the sodium ion ($N{a}^{+}$) pairs up with the bicarbonate ion ($HC{O}_3^{-}$) to form sodium bicarbonate ($NaHC{O}_3$).

So, the missing component C is sodium bicarbonate ($NaHC{O}_3$). The complete balanced equation is:

A Catch in the Chemistry: Why does this reaction proceed in the forward direction? The secret lies in solubility. In the cold aqueous mixture, sodium bicarbonate ($NaHC{O}_3$) is sparingly soluble compared to the other salts present. Due to the common ion effect and its inherently lower solubility at lower temperatures, the sodium bicarbonate precipitates out of the solution as solid crystals. These crystals can then be filtered out, dried, and heated to finally produce sodium carbonate.

By carefully analyzing the chemical logic of the Solvay process, we have successfully identified all the missing components.

We found that: A is $H_{2}O$ B is $C{O}_2$ C is $NaHC{O}_3$

This perfectly matches option (d). The beauty of this problem lies not just in balancing equations, but in understanding the industrial symphony of molecules interacting, exchanging ions, and precipitating to give us a product we use every single day.