cf₄ + br₂ → cbr₄ + f₂\ntotal # before (reactants) elements total # after (products)\nc\nf\nbr
cf₄ + br₂ → cbr₄ + f₂\ntotal # before (reactants) elements total # after (products)\nc\nf\nbr
Answer
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$CF_4$ + _$Br_2$ $\rightarrow$ 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$Br_2$ $\rightarrow$ 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$Br_2$ $\rightarrow$ 1$CBr_4$ + 2$F_2$
Now fill in the table:
| Total # Before (reactants) | Elements | Total # After (Products) |
|---|---|---|
| 1 | C | 1 |
| 4 | F | 4 |
| 4 | Br | 4 |
Answer:
1$CF_4$ + 2$Br_2$ $\rightarrow$ 1$CBr_4$ + 2$F_2$;
| Total # Before (reactants) | Elements | Total # After (Products) |
|---|---|---|
| 1 | C | 1 |
| 4 | F | 4 |
| 4 | Br | 4 |