This worksheet challenges students to balance ten different chemical equations, ranging from simple synthesis to complex double replacement reactions.

Here are the balanced chemical equations for each problem. I have worked through them step-by-step to ensure the number of atoms for each element is the same on both sides.

1. Iron and Sulfuric Acid
* Unbalanced: $\text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + \text{H}_2$
* Step 1: Balance Iron (Fe). There are 2 Fe on the right, so put a 2 in front of Fe on the left.
* $2\text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + \text{H}_2$
* Step 2: Balance Sulfate ($\text{SO}_4$). There are 3 sulfate groups on the right, so put a 3 in front of $\text{H}_2\text{SO}_4$.
* $2\text{Fe} + 3\text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + \text{H}_2$
* Step 3: Balance Hydrogen (H). There are now $3 \times 2 = 6$ hydrogens on the left. Put a 3 in front of $\text{H}_2$ on the right.
* $2\text{Fe} + 3\text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + 3\text{H}_2$

2. Methane Combustion
* Unbalanced: $\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$
* Step 1: Carbon (C) is already balanced (1 on each side).
* Step 2: Balance Hydrogen (H). There are 4 H on the left, so put a 2 in front of $\text{H}_2\text{O}$.
* $\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$
* Step 3: Balance Oxygen (O). On the right, there are 2 (from $\text{CO}_2$) + 2 (from water) = 4 oxygens. Put a 2 in front of $\text{O}_2$.
* $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$

3. Silicon Tetrachloride Hydrolysis
* Unbalanced: $\text{SiCl}_4 + \text{H}_2\text{O} \rightarrow \text{SiO}_2 + \text{HCl}$
* Step 1: Silicon (Si) is balanced.
* Step 2: Balance Chlorine (Cl). There are 4 Cl on the left, so put a 4 in front of HCl.
* $\text{SiCl}_4 + \text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}$
* Step 3: Balance Hydrogen (H). There are 4 H on the right, so put a 2 in front of $\text{H}_2\text{O}$. This also gives us 2 Oxygens, which matches the $\text{SiO}_2$.
* $\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}$

4. Silver Iodide and Sodium Sulfide
* Unbalanced: $\text{AgI} + \text{Na}_2\text{S} \rightarrow \text{Ag}_2\text{S} + \text{NaI}$
* Step 1: Balance Silver (Ag). There are 2 Ag on the right, so put a 2 in front of AgI.
* $2\text{AgI} + \text{Na}_2\text{S} \rightarrow \text{Ag}_2\text{S} + \text{NaI}$
* Step 2: Balance Iodine (I). There are 2 I on the left, so put a 2 in front of NaI. This also balances the Sodium (Na).
* $2\text{AgI} + \text{Na}_2\text{S} \rightarrow \text{Ag}_2\text{S} + 2\text{NaI}$

5. Ammonia Oxidation
* Unbalanced: $\text{NH}_3 + \text{O}_2 \rightarrow \text{NO} + \text{H}_2\text{O}$
* Step 1: Balance Hydrogen (H). Left has 3, Right has 2. The least common multiple is 6. Put a 2 in front of $\text{NH}_3$ and a 3 in front of $\text{H}_2\text{O}$.
* $2\text{NH}_3 + \text{O}_2 \rightarrow \text{NO} + 3\text{H}_2\text{O}$
* Step 2: Balance Nitrogen (N). Left has 2 N. Put a 2 in front of NO.
* $2\text{NH}_3 + \text{O}_2 \rightarrow 2\text{NO} + 3\text{H}_2\text{O}$
* Step 3: Balance Oxygen (O). Right has $2 + 3 = 5$ oxygens. We need 5 on the left, but $\text{O}_2$ comes in pairs. To fix this, double all coefficients from Step 2.
* $4\text{NH}_3 + \text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O}$
* Step 4: Now count Oxygen again. Right has $4 + 6 = 10$. Put a 5 in front of $\text{O}_2$.
* $4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O}$

6. Iron(III) Oxide Reduction
* Unbalanced: $\text{FeO}_3 + \text{CO} \rightarrow \text{Fe} + \text{CO}_2$
*(Note: Standard iron oxide is usually $\text{Fe}_2\text{O}_3$, but we will balance exactly what is written: $\text{FeO}_3$)*
* Step 1: Balance Oxygen in the iron compound. $\text{FeO}_3$ has 3 oxygens. We need 3 CO molecules to take them away and make 3 $\text{CO}_2$.
* $\text{FeO}_3 + 3\text{CO} \rightarrow \text{Fe} + 3\text{CO}_2$
* Step 2: Check Iron (Fe). 1 on left, 1 on right. It is balanced.
* $\text{FeO}_3 + 3\text{CO} \rightarrow \text{Fe} + 3\text{CO}_2$

7. Silicon Dioxide and Hydrofluoric Acid
* Unbalanced: $\text{SiO}_2 + \text{HF} \rightarrow \text{SiF}_4 + \text{H}_2\text{O}$
* Step 1: Balance Fluorine (F). There are 4 F on the right, so put a 4 in front of HF.
* $\text{SiO}_2 + 4\text{HF} \rightarrow \text{SiF}_4 + \text{H}_2\text{O}$
* Step 2: Balance Hydrogen (H). There are 4 H on the left, so put a 2 in front of $\text{H}_2\text{O}$. This also provides the 2 Oxygens needed for $\text{SiO}_2$.
* $\text{SiO}_2 + 4\text{HF} \rightarrow \text{SiF}_4 + 2\text{H}_2\text{O}$

8. Sodium Bromide and Chlorine
* Unbalanced: $\text{NaBr} + \text{Cl}_2 \rightarrow \text{NaCl} + \text{Br}_2$
* Step 1: Balance Chlorine (Cl) and Bromine (Br). Both appear as pairs on one side and singles on the other. Put a 2 in front of NaBr and a 2 in front of NaCl.
* $2\text{NaBr} + \text{Cl}_2 \rightarrow 2\text{NaCl} + \text{Br}_2$

9. Ammonium Phosphate and Lead(IV) Nitrate
* Unbalanced: $(\text{NH}_4)_3\text{PO}_4 + \text{Pb}(\text{NO}_3)_4 \rightarrow \text{Pb}_3(\text{PO}_4)_4 + \text{NH}_4\text{NO}_3$
* Step 1: Balance Lead (Pb). There are 3 Pb on the right, so put a 3 in front of $\text{Pb}(\text{NO}_3)_4$.
* $(\text{NH}_4)_3\text{PO}_4 + 3\text{Pb}(\text{NO}_3)_4 \rightarrow \text{Pb}_3(\text{PO}_4)_4 + \text{NH}_4\text{NO}_3$
* Step 2: Balance Phosphate ($\text{PO}_4$). There are 4 $\text{PO}_4$ groups on the right, so put a 4 in front of $(\text{NH}_4)_3\text{PO}_4$.
* $4(\text{NH}_4)_3\text{PO}_4 + 3\text{Pb}(\text{NO}_3)_4 \rightarrow \text{Pb}_3(\text{PO}_4)_4 + \text{NH}_4\text{NO}_3$
* Step 3: Balance Nitrogen and Ammonium.
* Left side: $4 \times 3 = 12$ Ammoniums ($\text{NH}_4$).
* Left side: $3 \times 4 = 12$ Nitrates ($\text{NO}_3$).
* So, put a 12 in front of $\text{NH}_4\text{NO}_3$.
* $4(\text{NH}_4)_3\text{PO}_4 + 3\text{Pb}(\text{NO}_3)_4 \rightarrow \text{Pb}_3(\text{PO}_4)_4 + 12\text{NH}_4\text{NO}_3$

10. Magnesium Hydroxide and Hydrochloric Acid
* Unbalanced: $\text{Mg(OH)}_2 + \text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2\text{O}$
* Step 1: Balance Chlorine (Cl). There are 2 Cl on the right, so put a 2 in front of HCl.
* $\text{Mg(OH)}_2 + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2\text{O}$
* Step 2: Balance Hydrogen (H) and Oxygen (O).
* Left: 2 H from hydroxide + 2 H from acid = 4 H total. 2 O from hydroxide.
* Right: Need 4 H and 2 O. Put a 2 in front of $\text{H}_2\text{O}$.
* $\text{Mg(OH)}_2 + 2\text{HCl} \rightarrow \text{MgCl}_2 + 2\text{H}_2\text{O}$

Final Answer:
1. 2 Fe + 3 H₂SO₄ → 1 Fe₂(SO₄)₃ + 3 H₂
2. 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
3. 1 SiCl₄ + 2 H₂O → 1 SiO₂ + 4 HCl
4. 2 AgI + 1 Na₂S → 1 Ag₂S + 2 NaI
5. 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O
6. 1 FeO₃ + 3 CO → 1 Fe + 3 CO₂
7. 1 SiO₂ + 4 HF → 1 SiF₄ + 2 H₂O
8. 2 NaBr + 1 Cl₂ → 2 NaCl + 1 Br₂
9. 4 (NH₄)₃PO₄ + 3 Pb(NO₃)₄ → 1 Pb₃(PO₄)₄ + 12 NH₄NO₃
10. 1 Mg(OH)₂ + 2 HCl → 1 MgCl₂ + 2 H₂O