# How would you balance the following equation? ?BF_3 + ?Li_2SO_3 -> ?B_2(SO_3)_3 + ?LiF

Dec 7, 2016

$2 B {F}_{3} + 3 L {i}_{2} S {O}_{3} \to {B}_{2} {\left(S O 3\right)}_{3} + 6 L i F$

#### Explanation:

Start with a species that you can see is not present in equal quantity on each side of the equation.

Perhaps you could recognize that Boron is not present in equal quantity on both sides of the equation - fix it by making the quantity equal eachother.

2BF_3+?Li_2SO_3 -> ?B_2(SO3)_3+?LiF

Notice that by doing this, we changed the number $F$.

Let's make the quantity of $F$ species equal.

2BF_3+?Li_2SO_3 -> ?B_2(SO3)_3+6LiF

You'll notice now that that we have changed the quantity of species $L i$ - let's fix that.

2BF_3+3Li_2SO_3 -> ?B_2(SO3)_3+6LiF

You'll notice now that all the species have equal quantities across the chemical system and that Boron Sulfite does not need a coefficient.

The balanced reaction is:

$2 B {F}_{3} + 3 L {i}_{2} S {O}_{3} \to {B}_{2} {\left(S O 3\right)}_{3} + 6 L i F$