How would you balance the following equation : $HI(g) \to {H_2}(g) + {I_2}(g)$.

Hint: See the number of atoms on both sides of the reaction, if they are not the same try to make the coefficient of each reactant or product change until you get them balanced. In the above equation you can multiply the molecule $HI$ to maintain the stoichiometry on the right hand side that is of ${H_2}$ and ${I_2}$.

Complete step by step answer:
In this equation we have hydrogen iodide $HI$ on the left hand side and hydrogen ${H_2}$ and iodine ${I_2}$ gases on the right. Now we know by seeing the equation that there are in total one hydrogen atom and one iodine atom on the left side of the equation. But the numbers are different on the right hand side of the equation, there are two hydrogens and two iodine on the right hand side.

So, by this information we can multiply the hydrogen iodide on the left hand side by any number which can make the atoms equal. If we multiply oppositely that means if we try to multiply the right hand side molecules they will increase the number of atoms and equal will make it hard to balance. So choosing first case of multiplying hydrogen iodide by a number $2$ by this the equation changes from $HI(g) \to {H_2}(g) + {I_2}(g)$ to $2\,HI(g) \to {H_2}(g) + {I_2}(g)$ .

Now let’s count the number of atoms on both sides, there are two hydrogens and two oxygen on the left of the equation and similarly there are two hydrogens and two oxygen on the right. So in this way we balance our equation so easily.

Note: Never multiply that side of an equation firstly which already contains more numeric value. In the above equation if we start balancing the equation by multiplying ${H_2}$ and ${I_2}$ with any number, it will gradually increase the amount of atoms at that side and the reaction gets more complex. So always start balancing by multiplying the molecule with a lesser numeric value.