Question

cf₄ + br₂ → cbr₄ + f₂
total # before (reactants) elements total # after (products)
c
f
br

Explanation:

Step1: Balance carbon atoms

Carbon is already balanced as there is 1 carbon atom on both sides of the equation. So for the coefficient of $CF_4$ and $CBr_4$, we can start with 1 for simplicity.
1$CF_4$ + _$Br_2$ $
ightarrow$ 1$CBr_4$ + _$F_2$

Step2: Balance bromine atoms

On the product - side, there are 4 bromine atoms in $CBr_4$. Since each $Br_2$ molecule has 2 bromine atoms, we need 2 $Br_2$ molecules to balance the bromine atoms.
1$CF_4$ + 2$Br_2$ $
ightarrow$ 1$CBr_4$ + _$F_2$

Step3: Balance fluorine atoms

On the reactant - side, there are 4 fluorine atoms in $CF_4$. On the product - side, we need 4 fluorine atoms in the form of $F_2$. Since each $F_2$ molecule has 2 fluorine atoms, we need 2 $F_2$ molecules.
1$CF_4$ + 2$Br_2$ $
ightarrow$ 1$CBr_4$ + 2$F_2$

Now fill in the table:

Total # Before (reactants)	Elements	Total # After (Products)
4	F	4
4	Br	4

Answer:

1$CF_4$ + 2$Br_2$ $
ightarrow$ 1$CBr_4$ + 2$F_2$;

Total # Before (reactants)	Elements	Total # After (Products)
4	F	4
4	Br	4