# How do you balance Al + F_2 -> AlF_3?

Feb 28, 2016

$A l + {F}_{2} \rightarrow A l {F}_{3}$
here F is unbalanced : on right hand side(product side) there are 3 F and on the left hand side there is 2F

use the L.C.M of the two numbers to balance on both sides
here L.C.M of (2,3)=6
so we try to get six F on both sides

$A l + 3 {F}_{2} \rightarrow 2 A l {F}_{3}$
now we need to balance Al
$2 A l + 3 {F}_{2} \rightarrow 2 A l {F}_{3}$

Feb 28, 2016

You balance it stoichiometrically: GARBAGE IN MUST EQUAL GARBAGE OUT.

#### Explanation:

A chemical reaction ALWAYS CONSERVES MASS AND CHARGE:

$2 A l \left(s\right) + 3 {F}_{2} \left(g\right) \rightarrow 2 A l {F}_{3} \left(s\right)$.

Is mass conserved in this representation? How? If you have to do calculations with this formula, to make it a little easier on yourself you could write:

$A l \left(s\right) + \frac{3}{2} {F}_{2} \left(g\right) \rightarrow A l {F}_{3} \left(s\right)$

What does $\frac{1}{2} {F}_{2} \left(g\right)$ represent in $\text{grams}$?