Question

How do you balance $Mn\,+\,HI\,\to \,{{H}_{2}}\,+\,Mn{{I}_{3}}$

Answer 1

The given chemical equation can be balanced with the help of hit and trial method. For this start balancing with the most complicated formula. Then balance all other atoms except $H$ and $O$. Then proceed to balance the $H\,and\,O$ atoms. Finally check whether the equation is balanced or not.

Complete step-by-step answer: Step 1: Start balancing with the most complex compound involved in the reaction.
Here the most complex compound is $Mn{{I}_{3}}$. Put a $1$ in front of $Mn{{I}_{3}}$, as shown below;
$Mn\,+\,HI\to \,{{H}_{2}}\,+\,1Mn{{I}_{3}}\,$
Step 2: Balance $Mn$
We have fixed one $Mn$ on the RHS, so in order to balance it we have to put $1$ in $Mn$ at LHS as shown below
$1Mn\,+\,HI\to \,{{H}_{2}}\,+\,1Mn{{I}_{3}}$
Step 3: Balance $I$
We have fixed $3\,I$ on the RHS, so we need to fix 3$I$ on the LHS. For this we will insert $3$ before $HI$as shown below
$1Mn\,+\,3HI\,\to \,{{H}_{2}}\,+\,1Mn{{I}_{3}}$
Step 4: Balance $H$
We have fixed $3H$ on the LHS, so we need 3H on the right. We have $2H$ on the RHS, so we must multiply ${{H}_{2}}$ with $\dfrac{3}{2}$ to balance the RHS as shown below
$1Mn\,+\,3HI\,\to \,\dfrac{3}{2}{{H}_{2}}\,+\,Mn{{I}_{3}}$
Now all the atoms are balanced, but while balancing we get a fraction on the RHS, in front of ${{H}_{2}}$
So, in order to remove a fraction, we will multiply the above balanced equation with $2$. On multiplying with $2$ we get the final balanced equation as shown below
$2Mn\,+\,6HI\to \,3{{H}_{2}}\,+\,Mn{{I}_{3}}$

Note: Once the reaction is balanced, cross check it by counting the number of atoms on RHS and LHS. If atoms both the sides are balanced, then the equation becomes balanced automatically. Always express the fraction in whole numbers while balancing a given reaction.