Practice Writing Chemical Equations from Word Equations - Free Printable

Here are the balanced chemical equations for the reactions listed in your worksheet. I have checked each one to ensure the number of atoms for every element is the same on both sides.

1. Zinc reacts with hydrogen chloride to form zinc chloride and hydrogen gas.
* Unbalanced: $\text{Zn} + \text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2$
* Balancing: You need 2 chlorine atoms and 2 hydrogen atoms on the left to match the right side.
* Equation: $\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2$

2. Sodium oxide reacts with water to form sodium hydroxide.
* Unbalanced: $\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow \text{NaOH}$
* Balancing: There are 2 sodiums on the left, so you need 2 NaOH on the right. This also balances the oxygen (2 total) and hydrogen (2 total).
* Equation: $\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH}$

3. Iron metal reacts with water to form $\text{Fe}_3\text{O}_4$ and hydrogen gas.
* Unbalanced: $\text{Fe} + \text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \text{H}_2$
* Balancing: Start with Iron. You need 3 Fe on the left. Now look at Oxygen. There are 4 O on the right, so you need 4 $\text{H}_2\text{O}$ on the left. This gives you 8 Hydrogens on the left, so you need 4 $\text{H}_2$ on the right.
* Equation: $3\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2$

4. Aluminum bromide reacts with chlorine gas to produce aluminum chloride and liquid bromine.
* *Note: Aluminum is Al, Bromine is Br, Chlorine is Cl. Bromine and Chlorine are diatomic ($\text{Br}_2$, $\text{Cl}_2$).*
* Unbalanced: $\text{AlBr}_3 + \text{Cl}_2 \rightarrow \text{AlCl}_3 + \text{Br}_2$
* Balancing: The trick here is balancing the odd/even numbers.
* To balance Bromine (3 vs 2), use a coefficient of 2 for $\text{AlBr}_3$ and 3 for $\text{Br}_2$.
* This gives 2 Aluminums, so put a 2 in front of $\text{AlCl}_3$.
* Now you have 6 Chlorines on the right ($2 \times 3$), so you need 3 $\text{Cl}_2$ on the left.
* Equation: $2\text{AlBr}_3 + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3 + 3\text{Br}_2$

5. Nitric acid ($\text{HNO}_3$) reacts with barium hydroxide to produce barium nitrate and water.
* *Note: Barium is Ba (charge +2), Nitrate is $\text{NO}_3$ (charge -1). Formula is $\text{Ba(NO}_3)_2$.*
* Unbalanced: $\text{HNO}_3 + \text{Ba(OH)}_2 \rightarrow \text{Ba(NO}_3)_2 + \text{H}_2\text{O}$
* Balancing: You need 2 nitrates on the left, so start with $2\text{HNO}_3$. This gives 2 Hydrogens from the acid and 2 Hydrogens from the base (total 4 H). This makes 2 waters ($2\text{H}_2\text{O}$).
* Equation: $2\text{HNO}_3 + \text{Ba(OH)}_2 \rightarrow \text{Ba(NO}_3)_2 + 2\text{H}_2\text{O}$

6. Calcium sulfite decomposes when heated to form calcium oxide and sulfur dioxide.
* *Note: Sulfite is $\text{SO}_3^{2-}$.*
* Unbalanced: $\text{CaSO}_3 \rightarrow \text{CaO} + \text{SO}_2$
* Balancing: Count the atoms. 1 Ca, 1 S, and 3 O on the left. On the right: 1 Ca, 1 S, and $1+2=3$ O. It is already balanced!
* Equation: $\text{CaSO}_3 \rightarrow \text{CaO} + \text{SO}_2$

7. Iron reacts with sulfuric acid ($\text{H}_2\text{SO}_4$) to form iron (II) sulfate and hydrogen gas.
* *Note: Iron (II) means $\text{Fe}^{2+}$. Sulfate is $\text{SO}_4^{2-}$. Formula is $\text{FeSO}_4$.*
* Unbalanced: $\text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + \text{H}_2$
* Balancing: 1 Fe, 2 H, 1 S, 4 O on both sides. It is already balanced.
* Equation: $\text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + \text{H}_2$

8. Ammonia ($\text{NH}_3$) burns in air to form nitrogen dioxide and water.
* Unbalanced: $\text{NH}_3 + \text{O}_2 \rightarrow \text{NO}_2 + \text{H}_2\text{O}$
* Balancing:
* Balance Hydrogen first: 3 on left, 2 on right. Least common multiple is 6. Put 2 before $\text{NH}_3$ and 3 before $\text{H}_2\text{O}$.
* Now Nitrogen: 2 N on left, so put 2 before $\text{NO}_2$.
* Now Oxygen: Right side has $(2 \times 2) + (3 \times 1) = 7$ oxygens. Left side has $\text{O}_2$. To get 7, you technically need 3.5 $\text{O}_2$.
* To remove the decimal, multiply the whole equation by 2.
* Equation: $4\text{NH}_3 + 7\text{O}_2 \rightarrow 4\text{NO}_2 + 6\text{H}_2\text{O}$

9. Carbon disulfide burns in air to form carbon dioxide and sulfur dioxide.
* *Note: Carbon disulfide is $\text{CS}_2$.*
* Unbalanced: $\text{CS}_2 + \text{O}_2 \rightarrow \text{CO}_2 + \text{SO}_2$
* Balancing:
* Carbon is balanced (1 each).
* Sulfur: 2 on left, so put 2 before $\text{SO}_2$.
* Oxygen: Right side has $2 (\text{from CO}_2) + 4 (\text{from } 2\text{SO}_2) = 6$ oxygens. So you need 3 $\text{O}_2$ on the left.
* Equation: $\text{CS}_2 + 3\text{O}_2 \rightarrow \text{CO}_2 + 2\text{SO}_2$

Final Answer:
1. $\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2$
2. $\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH}$
3. $3\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2$
4. $2\text{AlBr}_3 + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3 + 3\text{Br}_2$
5. $2\text{HNO}_3 + \text{Ba(OH)}_2 \rightarrow \text{Ba(NO}_3)_2 + 2\text{H}_2\text{O}$
6. $\text{CaSO}_3 \rightarrow \text{CaO} + \text{SO}_2$
7. $\text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + \text{H}_2$
8. $4\text{NH}_3 + 7\text{O}_2 \rightarrow 4\text{NO}_2 + 6\text{H}_2\text{O}$
9. $\text{CS}_2 + 3\text{O}_2 \rightarrow \text{CO}_2 + 2\text{SO}_2$