Balanced symbol equations - Higher tier
No atoms are created or destroyed in a chemical reaction – they are just rearranged.
Balanced symbol equations show the original arrangements and the new ones.
No atoms are created or destroyed when copper reacts with oxygen to form copper(II) oxide
Writing balanced symbol equations
| Step | Result |
| Check to see if there is an equal number of atoms of each element on both sides. There is not. | N2 + H2 → NH3 |
| There are two nitrogen atoms on the left but only one on the right, so put a big 2 on the left of the NH3. | N2 + H2 → 2NH3 |
| Check again. There are two hydrogen atoms on the left but (2 × 3) = 6 on the right, so put a big 3 in front of the H2. | N2 + 3H2 → 2NH3 |
| Check again to see if there are equal numbers of each element on both sides. There is. | (Two nitrogen atoms and six hydrogen atoms) |
| Add the state symbols if asked to do so. | N2(g) + 3H2(g) → 2NH3(g) |
| Step | Check to see if there is an equal number of atoms of each element on both sides. There is not. |
|---|---|
| Result | N2 + H2 → NH3 |
| Step | There are two nitrogen atoms on the left but only one on the right, so put a big 2 on the left of the NH3. |
|---|---|
| Result | N2 + H2 → 2NH3 |
| Step | Check again. There are two hydrogen atoms on the left but (2 × 3) = 6 on the right, so put a big 3 in front of the H2. |
|---|---|
| Result | N2 + 3H2 → 2NH3 |
| Step | Check again to see if there are equal numbers of each element on both sides. There is. |
|---|---|
| Result | (Two nitrogen atoms and six hydrogen atoms) |
| Step | Add the state symbols if asked to do so. |
|---|---|
| Result | N2(g) + 3H2(g) → 2NH3(g) |
