Question

How would you balance the following equation?
$?B{F_3} + ?L{i_2}S{O_3}\xrightarrow{{}}?{B_2}{(S{O_3})_3} + ?LiF$

Answer 1

Hint : Chemical equation should be balanced because there is a conservation of mass during the reaction. That means the number of atoms of a particular element before the reaction should be the same as the number of atoms after the reaction. While writing the equation, the number of atoms on the right hand side should be equal to the left hand side.

Complete Step By Step Answer:
The equation is $?B{F_3} + ?L{i_2}S{O_3}\xrightarrow{{}}?{B_2}{(S{O_3})_3} + ?LiF$
We see that there are 2 Boron compounds on the right side whereas there is just one Boron on the left side. So we multiply $B{F_3}$ with $2$ in order to neutralize Boron. So, we get:
$2B{F_3} + ?L{i_2}S{O_3}\xrightarrow{{}}{B_2}{(S{O_3})_3} + ?LiF$
Now, we see that there are $6$ Fluorine compounds on the left side and just $1$ Fluorine compounds on the right side. Hence, we multiply $LiF$ with $6$ .So, we get:
$2B{F_3} + ?L{i_2}S{O_3}\xrightarrow{{}}{B_2}{(S{O_3})_3} + 6LiF$
Now, we see that there are $6$ Lithium compounds on the right side and just $2$ Lithium compound on the left side. Hence, we multiply $L{i_2}(S{O_3})$ with $3$ .So, we get:
$2B{F_3} + 3L{i_2}S{O_3}\xrightarrow{{}}{B_2}{(S{O_3})_3} + 6LiF$
We can see that $S{O_3}$ is also balanced. So, the whole equation is balanced.
Hence, The balanced equation is $2B{F_3} + 3L{i_2}S{O_3}\xrightarrow{{}}{B_2}{(S{O_3})_3} + 6LiF$ .

Note :
If a chemical equation is unbalanced, then it indicates that the mass has either been created or destroyed during the reaction. But we know that this contradicts the law of conservation of mass which states that mass can neither be created and nor be destroyed. Hence, the chemical equations should always remain balanced. That's why we use these methods to balance the chemical equation.