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Question: How would you balance the following chemical equation? \(Hg{(N{O_3})_2}\, + KI \to \,Hg{I_2}\, + \...

How would you balance the following chemical equation?
Hg(NO3)2 +KI→ HgI2 + KNO3Hg{(N{O_3})_2}\, + KI \to \,Hg{I_2}\, + \,KN{O_3}

Explanation

Solution

The above equation forms when reactant between mercuric nitrate and potassium iodide takes place. So, while balancing a chemical equation we have to keep in mind that the number of atoms on each side should be equal. We can multiply the stoichiometric coefficient of each reactant and product and make each type of atom the same.

Complete step-by-step answer: The reaction takes place between mercuric nitrate Hg(NO3)2Hg{(N{O_3})_2} and potassium iodide KIKI which gives mercuric iodide HgI2 Hg{I_2}\, and potassium nitrate KNO3KN{O_3} as product. Now we have to make the number of atoms of each type equal. Thus firstly we have to count the number of atoms on both sides of the reaction. We have (1Hg, 3N, 6O, 1K and 1I)(1Hg,\,3N,\,6O,\,1K\,and\,1I) on the left hand side of reaction and (1Hg, 1N, 3O, 1K and 2I)(1Hg,\,1N,\,3O,\,1K\,and\,2I) on the right hand side of reaction.
Hg(NO3)2 +KI→ HgI2 + KNO3−−−−−−−(1)Hg{(N{O_3})_2}\, + KI \to \,Hg{I_2}\, + \,KN{O_3} - - - - - - - (1)
Let’s try to balance the equation by making changes of stoichiometry on the right hand side as there are fewer numbers of atoms. We can multiply the potassium nitrate on the right hand side by number 22 and potassium iodide also by number 22 . By this we get the unbalanced atom balanced. Now count the number of atoms on both sides to figure out the numbers.
Hg(NO3)2 +2KI→ HgI2 + 2KNO3−−−−−−−(2)Hg{(N{O_3})_2}\, + 2KI \to \,Hg{I_2}\, + \,2KN{O_3} - - - - - - - (2)
We have number of atoms of each type on left hand side of equation as (1Hg, 2N, 6O, 2K and 2I)(1Hg,\,2N,\,6O,\,2K\,and\,2I) and this is same as we have on right hand side (1Hg, 2N, 6O, 2K and 2I)(1Hg,\,2N,\,6O,\,2K\,and\,2I) thus, our equation is balanced.
Hg(NO3)2 +2KI→ HgI2 + 2KNO3−−−(Balanced chemical equation)Hg{(N{O_3})_2}\, + 2KI \to \,Hg{I_2}\, + \,2KN{O_3} - - - (Balanced\,chemical\,equation)

Note: As we see that the number of atoms on both sides becomes equal. There are some cases when the reaction is given with substances which have water of crystallization with them. So for those various reactions which occur with the water of crystallization we have to eliminate the water of crystallization and then repeat the same process of balancing. At last we end with the same number of atoms at last.