# How do you balance LiNO_3 + CaBr_2 -> Ca(NO_3)_2 + LiBr?

Mar 20, 2016

$2 L i N {O}_{3} + C a B {r}_{2} \to C a {\left(N {O}_{3}\right)}_{2} + 2 L i B r$

#### Explanation:

1. This is done through trial and error method, wherein coefficients are assigned to each compound temporarily and replaced until the no. of atoms of elements of both sides of the equation are balanced;
2. Generally, in this method you have to consider the complex compound first and work your way to the least one;
3. Polyatomic ions are usually consider as one unit, as in the case of nitrate ($N {O}_{3}$), which has 2 ions on the right side of the equation .
4. Looking at the left side $N {O}_{3}$ is only one, thus putting a coefficient of 2 would balance the nitrate and of course will change the no. of atoms of $L i$ which would increase to $2$.
5. Back to the right side, $L i$ is only one, thus putting again a coefficient of 2 would balance the no. of atoms of $L i$., and of course affect the number of atoms of $B r$ which would increase to $2$
6. Checking the left side for $B r$, the no. of atom is already 2, thus the equation is now balance.
7. You may create a T balance, to check whether elements have same number of atoms both sides of the equation;
$R e a c \tan t \top P r o \mathrm{du} c t$
$L i = 2$.......L$L i = 2$
$N = 2$........L$N = 2$
$O = 6$.........L$O = 6$
$C a = 1$.......L$C a = 1$
$B r = 2$.......L$B r = 2$