Question

How to balance chemical equations with hydrates?

Answer 1

Eliminate water of hydration firstly and then write the equation by forming the product. Balance it simply by making atoms of the same kind equal on both sides. At last write water of hydration as they are. This is because they are not actual, they are just written to express the presence of water in that particular compound.

Complete answer:
We can easily balance any chemical equation by making the number of atoms equal in left hand and right hand side. This equality can be maintained or we can say it can be achieved by multiplying the molecules with a proper coefficient.
But firstly let’s see what hydrates are and how their chemical equations form. This is because if we know about what hydrates are and what their chemical equation is, we can easily balance them.
Hydrates are substances that include water in their formulas. Like for example take copper sulphate, it has blue color which is due to the water of hydration it means there is some part of water in them by which they appeared blue. This water of hydration is not actually part of the chemical substance and this is reflected in the way the formula is written it means like this $CuS{O_4}.5{H_2}O$ .
It means that not all samples of CuSO4 are hydrates only those which have water of hydration means $5{H_2}O$ are called copper sulphate and CuSO4 without attached water molecules is referred to as anhydrous copper (II) sulphate.
There are many such compound which contains water of hydration like $FeS{O_4}.7{H_2}O$ ,
$MgS{O_4} \cdot {\mathbf{7}}{H_2}O\;$ , $ZnS{O_4}\; \cdot \;7{H_2}O$ and many more.
For solving these hydrates form a simple trick, keep in mind that this balancing is the same as all others, just solve the equation without considering the amount of water. If we eliminate water from hydration and then balance the equation and at last again write them as they are.
For example let’s take $CoC{l_2}\; \cdot \;6{H_2}O\;\, + \,N{a_2}S$
For completing this reaction as a balanced chemical equation, firstly exclude water of hydration $6{H_2}O\;$ then write a complete balanced equation as it is. $CoC{l_2}.6{H_2}O\;\, + \,N{a_2}S \to CoS + NaCl + \,{H_2}O$ .

Note: Don’t include water of hydration during balancing a chemical reaction of hydrates. If we start balancing the chemical equation by balancing hydrogens and oxygen they just make the answer wrong. The water of hydration is written to express the presence of water in that particular compound $CoC{l_2}.6{H_2}O$ .