Balancing Chemical Reactions interactive worksheet - Free Printable

To balance these chemical equations, we need to make sure that the number of atoms for each element is the same on both sides of the arrow (reactants and products). We do this by placing numbers (coefficients) in front of the chemical formulas. We cannot change the subscripts inside the formulas.

Let's go through them one by one:

1. Na + Cl₂ → NaCl
* Left side: 1 Na, 2 Cl
* Right side: 1 Na, 1 Cl
* To balance Chlorine, put a 2 in front of NaCl. Now we have 2 Na on the right.
* To balance Sodium, put a 2 in front of Na on the left.
* Balanced: 2 Na + 1 Cl₂ → 2 NaCl

2. Na + O₂ → Na₂O
* Left side: 1 Na, 2 O
* Right side: 2 Na, 1 O
* To balance Oxygen, put a 2 in front of Na₂O. Now we have 4 Na and 2 O on the right.
* To balance Sodium, put a 4 in front of Na on the left.
* Balanced: 4 Na + 1 O₂ → 2 Na₂O

3. Mg + O₂ → MgO
* Left side: 1 Mg, 2 O
* Right side: 1 Mg, 1 O
* To balance Oxygen, put a 2 in front of MgO. Now we have 2 Mg on the right.
* To balance Magnesium, put a 2 in front of Mg on the left.
* Balanced: 2 Mg + 1 O₂ → 2 MgO

4. Mg + HCl → MgCl₂ + H₂
* Left side: 1 Mg, 1 H, 1 Cl
* Right side: 1 Mg, 2 H, 2 Cl
* We need 2 Hydrogens and 2 Chlorines on the left. Put a 2 in front of HCl.
* Balanced: 1 Mg + 2 HCl → 1 MgCl₂ + 1 H₂

5. H₂O₂ → H₂ + O₂
* Left side: 2 H, 2 O
* Right side: 2 H, 2 O
* It is already balanced!
* Balanced: 1 H₂O₂ → 1 H₂ + 1 O₂

6. Al + HF → AlF₃ + H₂
* Left side: 1 Al, 1 H, 1 F
* Right side: 1 Al, 2 H, 3 F
* Look at Fluorine: Need 3 on left. Look at Hydrogen: Need 2 on left. The lowest common multiple for H and F here is tricky because they are in the same molecule (HF). Let's look at the product side H₂ (2 atoms) and AlF₃ (3 atoms).
* Try putting a 6 in front of HF. This gives 6 H and 6 F.
* On the right: To get 6 F, put a 2 in front of AlF₃. To get 6 H, put a 3 in front of H₂.
* Now check Aluminum: Right side has 2 Al (from 2AlF₃). Left needs 2 Al.
* Balanced: 2 Al + 6 HF → 2 AlF₃ + 3 H₂

7. NaClO₃ → NaCl + O₂
* Left side: 1 Na, 1 Cl, 3 O
* Right side: 1 Na, 1 Cl, 2 O
* Oxygen is 3 on left, 2 on right. Common multiple is 6.
* Put a 2 in front of NaClO₃ (gives 6 O).
* Put a 3 in front of O₂ (gives 6 O).
* Now we have 2 Na and 2 Cl on the left. Put a 2 in front of NaCl on the right.
* Balanced: 2 NaClO₃ → 2 NaCl + 3 O₂

8. ZnS + AlP → Zn₃P₂ + Al₂S₃
* Left side: 1 Zn, 1 S, 1 Al, 1 P
* Right side: 3 Zn, 2 P, 2 Al, 3 S
* Balance Zinc first: Put 3 in front of ZnS.
* Balance Sulfur: Left has 3 S (from 3ZnS), Right has 3 S (in Al₂S₃). Good.
* Balance Aluminum: Right has 2 Al. Put 2 in front of AlP.
* Balance Phosphorus: Left has 2 P (from 2AlP), Right has 2 P (in Zn₃P₂). Good.
* Balanced: 3 ZnS + 2 AlP → 1 Zn₃P₂ + 1 Al₂S₃

9. Ag₂S → Ag + S₈
* Left side: 2 Ag, 1 S
* Right side: 1 Ag, 8 S
* Balance Sulfur: Need 8 on left. Put 8 in front of Ag₂S.
* Now Silver: Left has 16 Ag (8 x 2). Right needs 16 Ag.
* Balanced: 8 Ag₂S → 16 Ag + 1 S₈

10. Ba(NO₃)₂ + H₂CO₃ → BaCO₃ + HNO₃
* Left side: 1 Ba, 2 NO₃ groups, 2 H, 1 CO₃ group
* Right side: 1 Ba, 1 CO₃ group, 1 H, 1 NO₃ group
* Nitrate (NO₃) is unbalanced (2 vs 1). Put 2 in front of HNO₃.
* Now Hydrogen is 2 on right, which matches the left (H₂CO₃).
* Everything else matches.
* Balanced: 1 Ba(NO₃)₂ + 1 H₂CO₃ → 1 BaCO₃ + 2 HNO₃

11. Pb(OH)₄ + Cu₂O → PbO₂ + CuOH
* Left side: 1 Pb, 4 OH groups, 2 Cu, 1 O (Wait, let's count atoms individually to be safe).
* Left: 1 Pb, 4 O (from OH), 4 H, 2 Cu, 1 O (from Cu₂O). Total O = 5.
* Right: 1 Pb, 2 O (from PbO₂), 1 Cu, 1 O, 1 H.
* Let's try balancing Copper first. Left has 2 Cu. Put 2 in front of CuOH.
* Now Right: 2 Cu, 2 H, 2 O (from CuOH) + 2 O (from PbO₂) = 4 O total.
* Left: Pb(OH)₄ has 4 O and 4 H. Cu₂O has 1 O. Total 5 O.
* Something is wrong with my atom counting or the equation logic. Let's re-examine the compounds.
* Reactants: Pb(OH)4 (1 Pb, 4 O, 4 H) + Cu2O (2 Cu, 1 O). Total: 1 Pb, 2 Cu, 5 O, 4 H.
* Products: PbO2 (1 Pb, 2 O) + CuOH (1 Cu, 1 O, 1 H).
* Let's try coefficient 4 for CuOH to match the 4 Hydrogens from Pb(OH)4.
* If 4 CuOH: Right side has 4 Cu, 4 O, 4 H. Plus PbO2 (1 Pb, 2 O). Total Right: 1 Pb, 4 Cu, 6 O, 4 H.
* Left side needs 4 Cu. So put 2 in front of Cu₂O.
* Left side now: 1 Pb, 4 O, 4 H (from Pb(OH)4) + 4 Cu, 2 O (from 2Cu2O). Total Left: 1 Pb, 4 Cu, 6 O, 4 H.
* Matches perfectly.
* Balanced: 1 Pb(OH)₄ + 2 Cu₂O → 1 PbO₂ + 4 CuOH

12. H₃PO₄ + HCl → PCl₅ + H₂O
* Left side: 3 H, 1 P, 4 O, 1 H, 1 Cl. Total H=4.
* Right side: 1 P, 5 Cl, 2 H, 1 O.
* Balance Chlorine: Need 5 on left. Put 5 in front of HCl.
* Now Hydrogen on left: 3 (from acid) + 5 (from HCl) = 8 H.
* Right side needs 8 H. Put 4 in front of H₂O.
* Check Oxygen: Left has 4 O. Right has 4 O (from 4H2O). Perfect.
* Balanced: 1 H₃PO₄ + 5 HCl → 1 PCl₅ + 4 H₂O

13. C₂H₆ + O₂ → CO₂ + H₂O
* Left: 2 C, 6 H. Right: 1 C, 2 H.
* Balance Carbon: Put 2 in front of CO₂.
* Balance Hydrogen: Put 3 in front of H₂O (3 x 2 = 6 H).
* Count Oxygen on right: (2 x 2) + (3 x 1) = 4 + 3 = 7 O.
* Left has O₂. We need 7 atoms, so we need 3.5 molecules. Multiply everything by 2 to clear the decimal.
* Start over with doubled coefficients:
* C₂H₆ becomes 2. CO₂ becomes 4. H₂O becomes 6.
* Check H: Left 2x6=12. Right 6x2=12. OK.
* Check C: Left 2x2=4. Right 4x1=4. OK.
* Check O: Right = (4x2) + (6x1) = 8 + 6 = 14.
* Left needs 14 O. Put 7 in front of O₂.
* Balanced: 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

14. C₃H₈ + O₂ → CO₂ + H₂O
* Left: 3 C, 8 H.
* Balance C: Put 3 in front of CO₂.
* Balance H: Put 4 in front of H₂O.
* Count O on right: (3 x 2) + (4 x 1) = 6 + 4 = 10.
* Left needs 10 O. Put 5 in front of O₂.
* Balanced: 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

15. C₅H₁₀ + O₂ → CO₂ + H₂O
* Left: 5 C, 10 H.
* Balance C: Put 5 in front of CO₂.
* Balance H: Put 5 in front of H₂O.
* Count O on right: (5 x 2) + (5 x 1) = 10 + 5 = 15.
* Left needs 15 O. This requires 7.5 O₂. Multiply all by 2.
* C₅H₁₀ becomes 2. CO₂ becomes 10. H₂O becomes 10.
* Recalculate O needed: (10 x 2) + (10 x 1) = 20 + 10 = 30.
* Left needs 30 O. Put 15 in front of O₂.
* Balanced: 2 C₅H₁₀ + 15 O₂ → 10 CO₂ + 10 H₂O

Final Answer:
1. 2, 1, 2
2. 4, 1, 2
3. 2, 1, 2
4. 1, 2, 1, 1
5. 1, 1, 1
6. 2, 6, 2, 3
7. 2, 2, 3
8. 3, 2, 1, 1
9. 8, 16, 1
10. 1, 1, 1, 2
11. 1, 2, 1, 4
12. 1, 5, 1, 4
13. 2, 7, 4, 6
14. 1, 5, 3, 4
15. 2, 15, 10, 10