Free Balancing Chemical Equations Worksheet Answer Key - Free Printable

The image provided is an Answer Key for a worksheet on balancing chemical equations. Since the answers are already filled in (shown in red), there is no unsolved problem to work through step-by-step.

However, to help you understand how these answers were reached, I will demonstrate the step-by-step balancing process for one of the more complex examples from the list: Equation #9.

Problem: Balance $C_8H_{18} + O_2 \rightarrow CO_2 + H_2O$

Step 1: Count the atoms on each side.
* Left Side (Reactants):
* Carbon (C): 8
* Hydrogen (H): 18
* Oxygen (O): 2
* Right Side (Products):
* Carbon (C): 1
* Hydrogen (H): 2
* Oxygen (O): 3 (2 in $CO_2$ + 1 in $H_2O$)

Step 2: Balance Carbon (C).
There are 8 Carbons on the left, so we need 8 on the right. We place an 8 in front of $CO_2$.
* Current equation: $1 C_8H_{18} + O_2 \rightarrow \mathbf{8} CO_2 + H_2O$

Step 3: Balance Hydrogen (H).
There are 18 Hydrogens on the left, so we need 18 on the right. Since water ($H_2O$) has 2 Hydrogens, we need 9 water molecules ($9 \times 2 = 18$). We place a 9 in front of $H_2O$.
* Current equation: $1 C_8H_{18} + O_2 \rightarrow 8 CO_2 + \mathbf{9} H_2O$

Step 4: Balance Oxygen (O).
Now count the total Oxygen on the right side:
* In $8 CO_2$: $8 \times 2 = 16$ Oxygen atoms.
* In $9 H_2O$: $9 \times 1 = 9$ Oxygen atoms.
* Total: $16 + 9 = 25$ Oxygen atoms.

On the left side, we have $O_2$. To get 25 atoms, we would technically need $12.5$ molecules of $O_2$ ($12.5 \times 2 = 25$).
* Temporary equation: $1 C_8H_{18} + \mathbf{12.5} O_2 \rightarrow 8 CO_2 + 9 H_2O$

Step 5: Remove decimals.
Chemical equations generally use whole numbers. To get rid of the $.5$, we multiply every coefficient in the entire equation by 2.
* $1 \times 2 = 2$
* $12.5 \times 2 = 25$
* $8 \times 2 = 16$
* $9 \times 2 = 18$

Final Check:
* Left: 16 C, 36 H, 50 O
* Right: 16 C, 36 H, 50 O ($32 + 18$)

The equation is balanced.

Final Answer:
The image contains the completed answer key for the worksheet. The balanced coefficients for the equations listed are:

1) 1 N₂ + 3 H₂ → 2 NH₃
2) 2 KClO₃ → 2 KCl + 3 O₂
3) 2 NaCl + 1 F₂ → 2 NaF + 1 Cl₂
4) 2 H₂ + 1 O₂ → 2 H₂O
5) 1 Pb(OH)₂ + 2 HCl → 2 H₂O + 1 PbCl₂
6) 2 AlBr₃ + 3 K₂SO₄ → 6 KBr + 1 Al₂(SO₄)₃
7) 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
8) 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
9) 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O
10) 1 FeCl₃ + 3 NaOH → 1 Fe(OH)₃ + 3 NaCl
11) 4 P + 5 O₂ → 2 P₂O₅
12) 2 Na + 2 H₂O → 2 NaOH + 1 H₂
13) 2 Ag₂O → 4 Ag + 1 O₂
14) 1 S₈ + 12 O₂ → 8 SO₃
15) 6 CO₂ + 6 H₂O → 1 C₆H₁₂O₆ + 6 O₂
16) 1 K + 1 MgBr → 1 KBr + 1 Mg *(Note: Standard magnesium bromide is MgBr₂, which would require different balancing, but this follows the key provided.)*
17) 2 HCl + 1 CaCO₃ → 1 CaCl₂ + 1 H₂O + 1 CO₂
18) 1 HNO + 1 NaHCO → 1 NaNO₃ + 1 H₂O + 1 CO₂
19) 2 H₂O + 1 O₂ → 2 H₂O₂
20) 2 NaBr + 1 CaF₂ → 2 NaF + 1 CaBr₂
21) 1 H₂SO₄ + 2 NaNO₂ → 2 HNO₂ + 1 Na₂SO₄