# How would you balance this chemical equation: 2Sb + I_2 -> 2SbI_3?

Apr 29, 2017

$2 S b + 3 {I}_{2} = = = 2 S b {I}_{3}$

#### Explanation:

There are six Iodines in the products so there must be six iodines in the reactants.

Iodine comes as diatomic gas molecule. ( as well as a grey crystal).

$3 \times {I}_{2} = 6 {I}^{-} 1$ therefore it requires 3 ${I}_{2}$ molecules to balance the equation

May 29, 2017

You follow a systematic procedure to balance the equation.

#### Explanation:

${\text{2Sb" + "I"_2 → "2SbI}}_{3}$

A method that often works is to balance everything other than $\text{O}$ and $\text{H}$ first, then balance $\text{O}$, and finally balance $\text{H}$.

Another useful procedure is to start with what looks like the most complicated formula.

The most complicated formula looks like ${\text{SbI}}_{3}$. It already has a 2 in front of it.

${\text{2Sb" + "I"_2 → color(red)(2)"SbI}}_{3}$

Balance $\text{Sb}$:

There are $\text{2 Sb}$ on the right, and there are $\text{2 Sb}$ on the right. $\text{Sb}$ is balanced.

$\textcolor{b l u e}{2} {\text{Sb" + "I"_2 → color(red)(2)"SbI}}_{3}$

Balance $\text{I}$:

We have fixed $\text{6 I}$ on the right, so we need $\text{6 I}$ on the left. Put a 3 in front of ${\text{I}}_{2}$.

$\textcolor{b l u e}{2} {\text{Sb" + color(orange)(3)"I"_2 → color(red)(2)"SbI}}_{3}$

Every formula now has a fixed coefficient. We should have a balanced equation.

Let’s check:

$\boldsymbol{\text{Atom"color(white)(m)"Left hand side"color(white)(m)"Right hand side}}$
$\textcolor{w h i t e}{m} \text{Sb} \textcolor{w h i t e}{m m m m m} 2 \textcolor{w h i t e}{m m m m m m m m} 2$
$\textcolor{w h i t e}{m} \text{I} \textcolor{w h i t e}{m m m m m m} 6 \textcolor{w h i t e}{m m m m m m m m} 6$

All atoms balance. The balanced equation is

$2 {\text{Sb" + 3"I"_2 → 2"SbI}}_{3}$