The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by NiCl

Q: How do I know the exact formula of gypsum?

Gypsum is a common chemical name you should memorize for JEE. Its formula is $CaSO_4 \cdot 2H_2O$. In this specific reaction, the water of crystallization comes from the acid-base neutralization process.

Q: What is the formula of the stable coordination compound formed?

Nickel(II) typically forms an octahedral complex with ammonia, which is a strong enough ligand to form $[Ni(NH_3)_6]^{2+}$. Since there are two chloride counter ions, the full formula is $[Ni(NH_3)_6]Cl_2$.

Q: Why did we assume all the ammonia is used up without checking for a limiting reagent?

The problem explicitly states that the ammonia is 'completely used' and both reactions are '100% complete'. Furthermore, if you calculate the moles, the $24 \text{ mol}$ of $NH_3$ and $4 \text{ mol}$ of Ni salt perfectly match the 6:1 stoichiometric ratio, meaning neither is in excess.

Q: Do I need to include the mass of water produced in the second reaction?

No, the question specifically asks for the combined weight of *gypsum* and the *nickel-ammonia coordination compound*. Water is a byproduct and should not be included in the final sum.

Q: How can I quickly calculate these large molar masses without making addition mistakes?

Break them down into familiar chunks. For example, the sulfate group ($SO_4$) is always $96 \text{ g/mol}$, ammonium ($NH_4$) is $18 \text{ g/mol}$, and water ($H_2O$) is $18 \text{ g/mol}$. Memorizing these common group masses saves time and significantly reduces calculation errors.

Question

The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by NiCl2.6H2O to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952g of NiCl2.6H2O are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is______.
(Atomic weights in g mol–1: H = 1, N = 14, O = 16, S = 32, Cl = 35.5, Ca = 40, Ni = 59)

Sigma Percentile · Accepted Answer

### The Beauty of Sequential Reactions

Imagine a perfectly choreographed dance of molecules, where the products of one reaction become the vital reactants for the next. This problem is a beautiful example of such a sequential process. We are tasked with tracking the flow of matter from a simple acid-base neutralization all the way to the formation of a complex coordination compound.

The key to mastering these problems is to never lose sight of the moles. Mass can be deceiving, but moles reveal the true stoichiometric ratios. Let's break this down step by step.

### Reaction 1: The Birth of Ammonia and Gypsum

Our journey begins with the preparation of ammonia. When ammonium sulphate, $(NH_4)_2SO_4$, is treated with calcium hydroxide, $Ca(OH)_2$, a classic double displacement and decomposition reaction occurs. The products are calcium sulphate dihydrate (commonly known as gypsum), ammonia gas, and water.

The balanced chemical equation is:
$$ (NH_4)_2SO_4 + Ca(OH)_2 ightarrow CaSO_4 \cdot 2H_2O + 2NH_3 $$

Notice the stoichiometry here. For every $1 	ext{ mole}$ of ammonium sulphate we consume, we produce $1 	ext{ mole}$ of gypsum and $2 	ext{ moles}$ of ammonia.

We are given $1584 	ext{ g}$ of ammonium sulphate. To find out how many moles this represents, we divide by its molar mass. The molar mass of $(NH_4)_2SO_4$ is calculated as $2(14 + 4) + 32 + 4(16) = 132 	ext{ g/mol}$.

$$ n_{(NH_4)_2SO_4} = \frac{1584 	ext{ g}}{132 	ext{ g/mol}} = 12 	ext{ moles} $$

Following our stoichiometric ratio, these $12 	ext{ moles}$ of reactant will yield exactly $12 	ext{ moles}$ of gypsum and $24 	ext{ moles}$ of ammonia gas.

### Reaction 2: Forging the Coordination Complex

Now, we take those $24 	ext{ moles}$ of ammonia and introduce them to a new reactant: nickel(II) chloride hexahydrate, $NiCl_2 \cdot 6H_2O$. Ammonia is a strong ligand, and it will readily displace the water molecules to form a stable octahedral coordination complex.

The balanced equation for this complexation is:
$$ NiCl_2 \cdot 6H_2O + 6NH_3 ightarrow [Ni(NH_3)_6]Cl_2 + 6H_2O $$

This equation tells us that $1 	ext{ mole}$ of the nickel salt requires exactly $6 	ext{ moles}$ of ammonia to fully coordinate.

Let's see how much nickel salt we actually have. We are given $952 	ext{ g}$ of $NiCl_2 \cdot 6H_2O$. Its molar mass is $59 + 2(35.5) + 6(18) = 238 	ext{ g/mol}$.

$$ n_{NiCl_2 \cdot 6H_2O} = \frac{952 	ext{ g}}{238 	ext{ g/mol}} = 4 	ext{ moles} $$

Here is where the magic happens. If we have $4 	ext{ moles}$ of the nickel salt, how much ammonia do we need? We need $4 	imes 6 = 24 	ext{ moles}$ of $NH_3$. And how much did we produce in the first reaction? Exactly $24 	ext{ moles}$!

This perfect match means there is no limiting reagent to worry about. Both reactants are completely consumed, yielding exactly $4 	ext{ moles}$ of the coordination compound, $[Ni(NH_3)_6]Cl_2$.

### The Final Weigh-In

We are in the home stretch. The problem asks for the combined weight of the gypsum produced in the first reaction and the coordination compound produced in the second.

First, we need their molar masses:
- Molar mass of Gypsum ($CaSO_4 \cdot 2H_2O$) = $40 + 32 + 64 + 36 = 172 	ext{ g/mol}$
- Molar mass of Complex ($[Ni(NH_3)_6]Cl_2$) = $59 + 6(17) + 71 = 232 	ext{ g/mol}$

Now, we simply multiply the moles we found earlier by these molar masses to get the final weights.

For Gypsum:
$$ W_{	ext{gypsum}} = 12 	ext{ moles} 	imes 172 	ext{ g/mol} = 2064 	ext{ g} $$

For the Coordination Complex:
$$ W_{	ext{complex}} = 4 	ext{ moles} 	imes 232 	ext{ g/mol} = 928 	ext{ g} $$

Finally, we add them together to find the grand total:
$$ W_{	ext{total}} = 2064 	ext{ g} + 928 	ext{ g} = \mathbf{2992 	ext{ g}} $$

And there we have it! By carefully tracking our moles through each step of the sequence, we arrive at the correct combined weight.

Enter Numerical Value:

Visualized Solution

Reaction Overview

Reaction 1: Preparation of Ammonia

Moles of Ammonium Sulphate

Products of Reaction 1

Reaction 2: Coordination Compound

Moles of Nickel Salt

Products of Reaction 2

Molar Masses of Products

Final Combined Weight

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Beauty of Sequential Reactions

Reaction 1

Reaction 2

The Final Weigh-In

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