# How do you balance "AgI + Fe"_2"(CO"_3)_3rarr"FeI"_3 + "Ag"_2"CO"_3"?

Sep 29, 2016

The balanced equation is $\textcolor{b l u e}{6} \text{AgI+Fe"_2"(CO"_3)_3}$$\rightarrow$$\textcolor{g r e e n}{2} {\text{FeI"_3+color(red)(3)"Ag"_2"CO}}_{3}$.

#### Explanation:

The law of the conservation of matter requires that the number of atoms of each element on each side of a chemical equation. This is done by adding coefficients in front of the formulas in the equation. A chemical formula should never be changed, only the amount.

$\text{AgI+Fe"_2"(CO"_3)_3}$$\rightarrow$${\text{FeI"_3+"Ag"_2"CO}}_{3}$

First notice that there is a polyatomic ion, the carbonate ion, $\text{CO"_3^(2-)}$ that occurs on both sides. When there is a polyatomic ion, it is treated as a single entity. So there are three carbonate ions on the left side and one on the right. So a coefficient of $3$ is needed in front of $\text{Ag"_2"CO"_3}$

$\text{AgI+Fe"_2"(CO"_3)_3}$$\rightarrow$${\text{FeI"_3+color(red)("3")"Ag"_2"CO}}_{3}$

Now there are six $\text{Ag}$ atoms on the right side, but one on the left. So a coefficient of $6$ needs to be added in front of the compound $\text{AgI}$.

$\textcolor{b l u e}{6} \text{AgI+Fe"_2"(CO"_3)_3}$$\rightarrow$${\text{FeI"_3+color(red)(3)"Ag"_2"CO}}_{3}$

Now there are six $\text{AgI}$, which gives us six $\text{I}$ atoms on the left side, but three on the right. So a coefficient of $2$ is needed in front of the compound $\text{FeI"_3}$.

$\textcolor{b l u e}{6} \text{AgI+Fe"_2"(CO"_3)_3}$$\rightarrow$$\textcolor{g r e e n}{2} {\text{FeI"_3+color(red)(3)"Ag"_2"CO}}_{3}$

This also balanced the $\text{Fe}$ atoms. So the balanced equation is

$\textcolor{b l u e}{6} \text{AgI+Fe"_2"(CO"_3)_3}$$\rightarrow$$\textcolor{g r e e n}{2} {\text{FeI"_3+color(red)(3)"Ag"_2"CO}}_{3}$