49 Balancing Chemical Equations Worksheets [with Answers] - Free Printable
![49 Balancing Chemical Equations Worksheets [with Answers]](/images2/chemistry-worksheet-balancing-equations/chemistry-worksheet-balancing-equations-3.jpg)
Educational worksheet: 49 Balancing Chemical Equations Worksheets [with Answers]. Download and print for classroom or home learning activities.
JPG
736×952
40.1 KB
Free · Personal Use
Quality Assured by Worksheets Library Team
Reviewed for educational accuracy and age-appropriateness
ID: #1671892
⭐
Show Answer Key & Explanations
Step-by-step solution for: 49 Balancing Chemical Equations Worksheets [with Answers]
▼
Show Answer Key & Explanations
Step-by-step solution for: 49 Balancing Chemical Equations Worksheets [with Answers]
To balance the chemical equations, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Here are the balanced equations for each problem:
---
- Start with the unbalanced equation:
\[
\text{N}_2 + \text{H}_2 \rightarrow \text{NH}_3
\]
- On the left side: 2 N and 2 H.
- On the right side: 1 N and 3 H in one molecule of NH₃.
- To balance nitrogen (N), we need 2 NH₃ molecules:
\[
\text{N}_2 + \text{H}_2 \rightarrow 2\text{NH}_3
\]
- Now, there are 6 H atoms on the right side, so we need 3 H₂ molecules on the left:
\[
\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3
\]
- Final balanced equation:
\[
\boxed{\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3}
\]
---
- Start with the unbalanced equation:
\[
\text{S}_8 + \text{O}_2 \rightarrow \text{SO}_3
\]
- On the left side: 8 S and 2 O.
- On the right side: 1 S and 3 O in one molecule of SO₃.
- To balance sulfur (S), we need 8 SO₃ molecules:
\[
\text{S}_8 + \text{O}_2 \rightarrow 8\text{SO}_3
\]
- Now, there are 24 O atoms on the right side, so we need 12 O₂ molecules on the left:
\[
\text{S}_8 + 12\text{O}_2 \rightarrow 8\text{SO}_3
\]
- Final balanced equation:
\[
\boxed{\text{S}_8 + 12\text{O}_2 \rightarrow 8\text{SO}_3}
\]
---
- Start with the unbalanced equation:
\[
\text{HgO} \rightarrow \text{Hg} + \text{O}_2
\]
- On the left side: 1 Hg and 1 O.
- On the right side: 1 Hg and 2 O in one molecule of O₂.
- To balance oxygen (O), we need 2 HgO molecules on the left:
\[
2\text{HgO} \rightarrow 2\text{Hg} + \text{O}_2
\]
- Final balanced equation:
\[
\boxed{2\text{HgO} \rightarrow 2\text{Hg} + \text{O}_2}
\]
---
- Start with the unbalanced equation:
\[
\text{Zn} + \text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2
\]
- On the left side: 1 Zn, 1 H, and 1 Cl.
- On the right side: 1 Zn, 2 Cl in ZnCl₂, and 2 H in H₂.
- To balance chlorine (Cl), we need 2 HCl molecules on the left:
\[
\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2
\]
- Final balanced equation:
\[
\boxed{\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2}
\]
---
- Start with the unbalanced equation:
\[
\text{SiCl}_4 + \text{H}_2\text{O} \rightarrow \text{H}_4\text{SiO}_4 + \text{HCl}
\]
- On the left side: 1 Si, 4 Cl, 2 H, and 1 O.
- On the right side: 1 Si, 4 H, 1 O, and 1 Cl in H₄SiO₄, and 1 H and 1 Cl in HCl.
- To balance chlorine (Cl), we need 4 HCl molecules on the right:
\[
\text{SiCl}_4 + \text{H}_2\text{O} \rightarrow \text{H}_4\text{SiO}_4 + 4\text{HCl}
\]
- Now, there are 8 H atoms on the right side, so we need 4 H₂O molecules on the left:
\[
\text{SiCl}_4 + 4\text{H}_2\text{O} \rightarrow \text{H}_4\text{SiO}_4 + 4\text{HCl}
\]
- Final balanced equation:
\[
\boxed{\text{SiCl}_4 + 4\text{H}_2\text{O} \rightarrow \text{H}_4\text{SiO}_4 + 4\text{HCl}}
\]
---
- Start with the unbalanced equation:
\[
\text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2
\]
- On the left side: 1 Na, 2 H, and 1 O.
- On the right side: 1 Na, 1 O, 1 H in NaOH, and 2 H in H₂.
- To balance sodium (Na), we need 2 NaOH molecules on the right:
\[
\text{Na} + \text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2
\]
- Now, there are 4 H atoms on the right side, so we need 2 H₂O molecules on the left:
\[
2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2
\]
- Final balanced equation:
\[
\boxed{2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2}
\]
---
- Start with the unbalanced equation:
\[
\text{H}_3\text{PO}_4 \rightarrow \text{H}_4\text{P}_2\text{O}_7 + \text{H}_2\text{O}
\]
- On the left side: 3 H, 1 P, and 4 O.
- On the right side: 4 H, 2 P, and 7 O in H₄P₂O₇, and 2 H and 1 O in H₂O.
- To balance phosphorus (P), we need 2 H₃PO₄ molecules on the left:
\[
2\text{H}_3\text{PO}_4 \rightarrow \text{H}_4\text{P}_2\text{O}_7 + \text{H}_2\text{O}
\]
- Now, there are 6 H atoms on the left side, so we need 1 H₂O molecule on the right:
\[
2\text{H}_3\text{PO}_4 \rightarrow \text{H}_4\text{P}_2\text{O}_7 + \text{H}_2\text{O}
\]
- Final balanced equation:
\[
\boxed{2\text{H}_3\text{PO}_4 \rightarrow \text{H}_4\text{P}_2\text{O}_7 + \text{H}_2\text{O}}
\]
---
- Start with the unbalanced equation:
\[
\text{Si}_2\text{H}_3 + \text{O}_2 \rightarrow \text{SiO}_2 + \text{H}_2\text{O}
\]
- On the left side: 2 Si, 3 H, and 2 O.
- On the right side: 2 Si, 2 O in SiO₂, and 2 H and 1 O in H₂O.
- To balance silicon (Si), we need 1 SiO₂ molecule for each Si atom:
\[
\text{Si}_2\text{H}_3 + \text{O}_2 \rightarrow 2\text{SiO}_2 + \text{H}_2\text{O}
\]
- Now, there are 4 O atoms on the right side from SiO₂, so we need 3 O atoms from H₂O:
\[
\text{Si}_2\text{H}_3 + \text{O}_2 \rightarrow 2\text{SiO}_2 + 3\text{H}_2\text{O}
\]
- Now, there are 6 H atoms on the right side, so we need 3 H₂O molecules on the right:
\[
\text{Si}_2\text{H}_3 + \text{O}_2 \rightarrow 2\text{SiO}_2 + 3\text{H}_2\text{O}
\]
- Finally, balance oxygen (O):
\[
\text{Si}_2\text{H}_3 + \frac{7}{2}\text{O}_2 \rightarrow 2\text{SiO}_2 + 3\text{H}_2\text{O}
\]
Multiply through by 2 to eliminate the fraction:
\[
2\text{Si}_2\text{H}_3 + 7\text{O}_2 \rightarrow 4\text{SiO}_2 + 6\text{H}_2\text{O}
\]
- Final balanced equation:
\[
\boxed{2\text{Si}_2\text{H}_3 + 7\text{O}_2 \rightarrow 4\text{SiO}_2 + 6\text{H}_2\text{O}}
\]
---
- Start with the unbalanced equation:
\[
\text{Al(OH)}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O}
\]
- On the left side: 1 Al, 3 O, 3 H, 1 S, and 4 O.
- On the right side: 2 Al, 3 S, 12 O in Al₂(SO₄)₃, and 2 H and 1 O in H₂O.
- To balance aluminum (Al), we need 2 Al(OH)₃ molecules on the left:
\[
2\text{Al(OH)}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O}
\]
- Now, there are 6 O atoms from Al(OH)₃ and 4 O atoms from H₂SO₄, so we need 13 O atoms on the right:
\[
2\text{Al(OH)}_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6\text{H}_2\text{O}
\]
- Final balanced equation:
\[
\boxed{2\text{Al(OH)}_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6\text{H}_2\text{O}}
\]
---
- Start with the unbalanced equation:
\[
\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3
\]
- On the left side: 1 Fe and 2 O.
- On the right side: 2 Fe and 3 O in Fe₂O₃.
- To balance iron (Fe), we need 2 Fe atoms on the left:
\[
2\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3
\]
- Now, there are 3 O atoms on the right side, so we need 1.5 O₂ molecules on the left:
\[
2\text{Fe} + \frac{3}{2}\text{O}_2 \rightarrow \text{Fe}_2\text{O}_3
\]
Multiply through by 2 to eliminate the fraction:
\[
4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3
\]
- Final balanced equation:
\[
\boxed{4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3}
\]
---
- Start with the unbalanced equation:
\[
\text{Fe}_2(\text{SO}_4)_3 + \text{KOH} \rightarrow \text{K}_2\text{SO}_4 + \text{Fe(OH)}_3
\]
- On the left side: 2 Fe, 3 S, 12 O, 1 K, and 1 H.
- On the right side: 2 K, 1 S, 7 O in K₂SO₄, and 1 Fe, 3 O, and 3 H in Fe(OH)₃.
- To balance iron (Fe), we need 2 Fe(OH)₃ molecules on the right:
\[
\text{Fe}_2(\text{SO}_4)_3 + \text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{Fe(OH)}_3
\]
- Now, there are 2 K atoms on the right side, so we need 2 KOH molecules on the left:
\[
\text{Fe}_2(\text{SO}_4)_3 + 6\text{KOH} \rightarrow 3\text{K}_2\text{SO}_4 + 2\text{Fe(OH)}_3
\]
- Final balanced equation:
\[
\boxed{\text{Fe}_2(\text{SO}_4)_3 + 6\text{KOH} \rightarrow 3\text{K}_2\text{SO}_4 + 2\text{Fe(OH)}_3}
\]
---
- Start with the unbalanced equation:
\[
\text{FeS}_2 + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 + \text{SO}_2
\]
- On the left side: 1 Fe, 2 S, and 2 O.
- On the right side: 2 Fe, 3 O in Fe₂O₃, and 1 S and 2 O in SO₂.
- To balance iron (Fe), we need 2 FeS₂ molecules on the left:
\[
2\text{FeS}_2 + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 + \text{SO}_2
\]
- Now, there are 4 S atoms on the left side, so we need 4 SO₂ molecules on the right:
\[
2\text{FeS}_2 + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 + 4\text{SO}_2
\]
- Now, there are 8 O atoms on the right side, so we need 11/2 O₂ molecules on the left:
\[
2\text{FeS}_2 + \frac{11}{2}\text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 + 4\text{SO}_2
\]
Multiply through by 2 to eliminate the fraction:
\[
4\text{FeS}_2 + 11\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2
\]
- Final balanced equation:
\[
\boxed{4\text{FeS}_2 + 11\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2}
\]
---
- Start with the unbalanced equation:
\[
\text{Al} + \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}
\]
- On the left side: 1 Al, 1 Fe, and 1 O.
- On the right side: 2 Al, 3 O in Al₂O₃, and 1 Fe.
- To balance aluminum (Al), we need 2 Al atoms on the left:
\[
2\text{Al} + \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}
\]
- Now, there are 3 O atoms on the right side, so we need 3 FeO molecules on the left:
\[
2\text{Al} + 3\text{FeO} \rightarrow \text{Al}_2\text{O}_3 + 3\text{Fe}
\]
- Final balanced equation:
\[
\boxed{2\text{Al} + 3\text{FeO} \rightarrow \text{Al}_2\text{O}_3 + 3\text{Fe}}
\]
---
- Start with the unbalanced equation:
\[
\text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2
\]
- On the left side: 2 Na, 1 C, 3 O, 1 H, and 1 Cl.
- On the right side: 1 Na, 1 Cl in NaCl, 2 H and 1 O in H₂O, and 1 C and 2 O in CO₂.
- To balance sodium (Na), we need 2 NaCl molecules on the right:
\[
\text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2
\]
- Now, there are 2 Cl atoms on the right side, so we need 2 HCl molecules on the left:
\[
\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2
\]
- Final balanced equation:
\[
\boxed{\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2}
\]
---
- Start with the unbalanced equation:
\[
\text{K} + \text{Br}_2 \rightarrow \text{KBr}
\]
- On the left side: 1 K and 2 Br.
- On the right side: 1 K and 1 Br in KBr.
- To balance bromine (Br), we need 2 KBr molecules on the right:
\[
\text{K} + \text{Br}_2 \rightarrow 2\text{KBr}
\]
- Now, there are 2 K atoms on the right side, so we need 2 K atoms on the left:
\[
2\text{K} + \text{Br}_2 \rightarrow 2\text{KBr}
\]
- Final balanced equation:
\[
\boxed{2\text{K} + \text{Br}_2 \rightarrow 2\text{KBr}}
\]
---
- Start with the unbalanced equation:
\[
\text{P}_4 + \text{O}_2 \rightarrow \text{P}_2\text{O}_5
\]
- On the left side: 4 P and 2 O.
- On the right side: 2 P and 5 O in P₂O₅.
- To balance phosphorus (P), we need 2 P₂O₅ molecules on the right:
\[
\text{P}_4 + \text{O}_2 \rightarrow 2\text{P}_2\text{O}_5
\]
- Now, there are 10 O atoms on the right side, so we need 5 O₂ molecules on the left:
\[
\text{P}_4 + 5\text{O}_2 \rightarrow 2\text{P}_2\text{O}_5
\]
- Final balanced equation:
\[
\boxed{\text{P}_4 + 5\text{O}_2 \rightarrow 2\text{P}_2\text{O}_5}
\]
---
- Start with the unbalanced equation:
\[
\text{C}_2\text{H}_2 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
\]
- On the left side: 2 C, 2 H, and 2 O.
- On the right side: 1 C, 2 O in CO₂, and 2 H and 1 O in H₂O.
- To balance carbon (C), we need 2 CO₂ molecules on the right:
\[
\text{C}_2\text{H}_2 + \text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O}
\]
- Now, there are 4 O atoms on the right side from CO₂ and 1 O atom from H₂O, so we need 5/2 O₂ molecules on the left:
\[
\text{C}_2\text{H}_2 + \frac{5}{2}\text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O}
\]
Multiply through by 2 to eliminate the fraction:
\[
2\text{C}_2\text{H}_2 + 5\text{O}_2 \rightarrow 4\text{CO}_2 + 2\text{H}_2\text{O}
\]
- Final balanced equation:
\[
\boxed{2\text{C}_2\text{H}_2 + 5\text{O}_2 \rightarrow 4\text{CO}_2 + 2\text{H}_2\text{O}}
\]
---
\[
\boxed{
\begin{aligned}
1. & \quad \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \\
2. & \quad \text{S}_8 + 12\text{O}_2 \rightarrow 8\text{SO}_3 \\
3. & \quad 2\text{HgO} \rightarrow 2\text{Hg} + \text{O}_2 \\
4. & \quad \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \\
5. & \quad \text{SiCl}_4 + 4\text{H}_2\text{O} \rightarrow \text{H}_4\text{SiO}_4 + 4\text{HCl} \\
6. & \quad 2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 \\
7. & \quad 2\text{H}_3\text{PO}_4 \rightarrow \text{H}_4\text{P}_2\text{O}_7 + \text{H}_2\text{O} \\
8. & \quad 2\text{Si}_2\text{H}_3 + 7\text{O}_2 \rightarrow 4\text{SiO}_2 + 6\text{H}_2\text{O} \\
9. & \quad 2\text{Al(OH)}_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6\text{H}_2\text{O} \\
10. & \quad 4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 \\
11. & \quad \text{Fe}_2(\text{SO}_4)_3 + 6\text{KOH} \rightarrow 3\text{K}_2\text{SO}_4 + 2\text{Fe(OH)}_3 \\
12. & \quad 4\text{FeS}_2 + 11\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2 \\
13. & \quad 2\text{Al} + 3\text{FeO} \rightarrow \text{Al}_2\text{O}_3 + 3\text{Fe} \\
14. & \quad \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \\
15. & \quad 2\text{K} + \text{Br}_2 \rightarrow 2\text{KBr} \\
16. & \quad \text{P}_4 + 5\text{O}_2 \rightarrow 2\text{P}_2\text{O}_5 \\
17. & \quad 2\text{C}_2\text{H}_2 + 5\text{O}_2 \rightarrow 4\text{CO}_2 + 2\text{H}_2\text{O}
\end{aligned}
}
\]
---
1. \( \text{N}_2 + \text{H}_2 \rightarrow \text{NH}_3 \)
- Start with the unbalanced equation:
\[
\text{N}_2 + \text{H}_2 \rightarrow \text{NH}_3
\]
- On the left side: 2 N and 2 H.
- On the right side: 1 N and 3 H in one molecule of NH₃.
- To balance nitrogen (N), we need 2 NH₃ molecules:
\[
\text{N}_2 + \text{H}_2 \rightarrow 2\text{NH}_3
\]
- Now, there are 6 H atoms on the right side, so we need 3 H₂ molecules on the left:
\[
\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3
\]
- Final balanced equation:
\[
\boxed{\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3}
\]
---
2. \( \text{S}_8 + \text{O}_2 \rightarrow \text{SO}_3 \)
- Start with the unbalanced equation:
\[
\text{S}_8 + \text{O}_2 \rightarrow \text{SO}_3
\]
- On the left side: 8 S and 2 O.
- On the right side: 1 S and 3 O in one molecule of SO₃.
- To balance sulfur (S), we need 8 SO₃ molecules:
\[
\text{S}_8 + \text{O}_2 \rightarrow 8\text{SO}_3
\]
- Now, there are 24 O atoms on the right side, so we need 12 O₂ molecules on the left:
\[
\text{S}_8 + 12\text{O}_2 \rightarrow 8\text{SO}_3
\]
- Final balanced equation:
\[
\boxed{\text{S}_8 + 12\text{O}_2 \rightarrow 8\text{SO}_3}
\]
---
3. \( \text{HgO} \rightarrow \text{Hg} + \text{O}_2 \)
- Start with the unbalanced equation:
\[
\text{HgO} \rightarrow \text{Hg} + \text{O}_2
\]
- On the left side: 1 Hg and 1 O.
- On the right side: 1 Hg and 2 O in one molecule of O₂.
- To balance oxygen (O), we need 2 HgO molecules on the left:
\[
2\text{HgO} \rightarrow 2\text{Hg} + \text{O}_2
\]
- Final balanced equation:
\[
\boxed{2\text{HgO} \rightarrow 2\text{Hg} + \text{O}_2}
\]
---
4. \( \text{Zn} + \text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \)
- Start with the unbalanced equation:
\[
\text{Zn} + \text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2
\]
- On the left side: 1 Zn, 1 H, and 1 Cl.
- On the right side: 1 Zn, 2 Cl in ZnCl₂, and 2 H in H₂.
- To balance chlorine (Cl), we need 2 HCl molecules on the left:
\[
\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2
\]
- Final balanced equation:
\[
\boxed{\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2}
\]
---
5. \( \text{SiCl}_4 + \text{H}_2\text{O} \rightarrow \text{H}_4\text{SiO}_4 + \text{HCl} \)
- Start with the unbalanced equation:
\[
\text{SiCl}_4 + \text{H}_2\text{O} \rightarrow \text{H}_4\text{SiO}_4 + \text{HCl}
\]
- On the left side: 1 Si, 4 Cl, 2 H, and 1 O.
- On the right side: 1 Si, 4 H, 1 O, and 1 Cl in H₄SiO₄, and 1 H and 1 Cl in HCl.
- To balance chlorine (Cl), we need 4 HCl molecules on the right:
\[
\text{SiCl}_4 + \text{H}_2\text{O} \rightarrow \text{H}_4\text{SiO}_4 + 4\text{HCl}
\]
- Now, there are 8 H atoms on the right side, so we need 4 H₂O molecules on the left:
\[
\text{SiCl}_4 + 4\text{H}_2\text{O} \rightarrow \text{H}_4\text{SiO}_4 + 4\text{HCl}
\]
- Final balanced equation:
\[
\boxed{\text{SiCl}_4 + 4\text{H}_2\text{O} \rightarrow \text{H}_4\text{SiO}_4 + 4\text{HCl}}
\]
---
6. \( \text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2 \)
- Start with the unbalanced equation:
\[
\text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2
\]
- On the left side: 1 Na, 2 H, and 1 O.
- On the right side: 1 Na, 1 O, 1 H in NaOH, and 2 H in H₂.
- To balance sodium (Na), we need 2 NaOH molecules on the right:
\[
\text{Na} + \text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2
\]
- Now, there are 4 H atoms on the right side, so we need 2 H₂O molecules on the left:
\[
2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2
\]
- Final balanced equation:
\[
\boxed{2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2}
\]
---
7. \( \text{H}_3\text{PO}_4 \rightarrow \text{H}_4\text{P}_2\text{O}_7 + \text{H}_2\text{O} \)
- Start with the unbalanced equation:
\[
\text{H}_3\text{PO}_4 \rightarrow \text{H}_4\text{P}_2\text{O}_7 + \text{H}_2\text{O}
\]
- On the left side: 3 H, 1 P, and 4 O.
- On the right side: 4 H, 2 P, and 7 O in H₄P₂O₇, and 2 H and 1 O in H₂O.
- To balance phosphorus (P), we need 2 H₃PO₄ molecules on the left:
\[
2\text{H}_3\text{PO}_4 \rightarrow \text{H}_4\text{P}_2\text{O}_7 + \text{H}_2\text{O}
\]
- Now, there are 6 H atoms on the left side, so we need 1 H₂O molecule on the right:
\[
2\text{H}_3\text{PO}_4 \rightarrow \text{H}_4\text{P}_2\text{O}_7 + \text{H}_2\text{O}
\]
- Final balanced equation:
\[
\boxed{2\text{H}_3\text{PO}_4 \rightarrow \text{H}_4\text{P}_2\text{O}_7 + \text{H}_2\text{O}}
\]
---
8. \( \text{Si}_2\text{H}_3 + \text{O}_2 \rightarrow \text{SiO}_2 + \text{H}_2\text{O} \)
- Start with the unbalanced equation:
\[
\text{Si}_2\text{H}_3 + \text{O}_2 \rightarrow \text{SiO}_2 + \text{H}_2\text{O}
\]
- On the left side: 2 Si, 3 H, and 2 O.
- On the right side: 2 Si, 2 O in SiO₂, and 2 H and 1 O in H₂O.
- To balance silicon (Si), we need 1 SiO₂ molecule for each Si atom:
\[
\text{Si}_2\text{H}_3 + \text{O}_2 \rightarrow 2\text{SiO}_2 + \text{H}_2\text{O}
\]
- Now, there are 4 O atoms on the right side from SiO₂, so we need 3 O atoms from H₂O:
\[
\text{Si}_2\text{H}_3 + \text{O}_2 \rightarrow 2\text{SiO}_2 + 3\text{H}_2\text{O}
\]
- Now, there are 6 H atoms on the right side, so we need 3 H₂O molecules on the right:
\[
\text{Si}_2\text{H}_3 + \text{O}_2 \rightarrow 2\text{SiO}_2 + 3\text{H}_2\text{O}
\]
- Finally, balance oxygen (O):
\[
\text{Si}_2\text{H}_3 + \frac{7}{2}\text{O}_2 \rightarrow 2\text{SiO}_2 + 3\text{H}_2\text{O}
\]
Multiply through by 2 to eliminate the fraction:
\[
2\text{Si}_2\text{H}_3 + 7\text{O}_2 \rightarrow 4\text{SiO}_2 + 6\text{H}_2\text{O}
\]
- Final balanced equation:
\[
\boxed{2\text{Si}_2\text{H}_3 + 7\text{O}_2 \rightarrow 4\text{SiO}_2 + 6\text{H}_2\text{O}}
\]
---
9. \( \text{Al(OH)}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O} \)
- Start with the unbalanced equation:
\[
\text{Al(OH)}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O}
\]
- On the left side: 1 Al, 3 O, 3 H, 1 S, and 4 O.
- On the right side: 2 Al, 3 S, 12 O in Al₂(SO₄)₃, and 2 H and 1 O in H₂O.
- To balance aluminum (Al), we need 2 Al(OH)₃ molecules on the left:
\[
2\text{Al(OH)}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O}
\]
- Now, there are 6 O atoms from Al(OH)₃ and 4 O atoms from H₂SO₄, so we need 13 O atoms on the right:
\[
2\text{Al(OH)}_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6\text{H}_2\text{O}
\]
- Final balanced equation:
\[
\boxed{2\text{Al(OH)}_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6\text{H}_2\text{O}}
\]
---
10. \( \text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 \)
- Start with the unbalanced equation:
\[
\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3
\]
- On the left side: 1 Fe and 2 O.
- On the right side: 2 Fe and 3 O in Fe₂O₃.
- To balance iron (Fe), we need 2 Fe atoms on the left:
\[
2\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3
\]
- Now, there are 3 O atoms on the right side, so we need 1.5 O₂ molecules on the left:
\[
2\text{Fe} + \frac{3}{2}\text{O}_2 \rightarrow \text{Fe}_2\text{O}_3
\]
Multiply through by 2 to eliminate the fraction:
\[
4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3
\]
- Final balanced equation:
\[
\boxed{4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3}
\]
---
11. \( \text{Fe}_2(\text{SO}_4)_3 + \text{KOH} \rightarrow \text{K}_2\text{SO}_4 + \text{Fe(OH)}_3 \)
- Start with the unbalanced equation:
\[
\text{Fe}_2(\text{SO}_4)_3 + \text{KOH} \rightarrow \text{K}_2\text{SO}_4 + \text{Fe(OH)}_3
\]
- On the left side: 2 Fe, 3 S, 12 O, 1 K, and 1 H.
- On the right side: 2 K, 1 S, 7 O in K₂SO₄, and 1 Fe, 3 O, and 3 H in Fe(OH)₃.
- To balance iron (Fe), we need 2 Fe(OH)₃ molecules on the right:
\[
\text{Fe}_2(\text{SO}_4)_3 + \text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{Fe(OH)}_3
\]
- Now, there are 2 K atoms on the right side, so we need 2 KOH molecules on the left:
\[
\text{Fe}_2(\text{SO}_4)_3 + 6\text{KOH} \rightarrow 3\text{K}_2\text{SO}_4 + 2\text{Fe(OH)}_3
\]
- Final balanced equation:
\[
\boxed{\text{Fe}_2(\text{SO}_4)_3 + 6\text{KOH} \rightarrow 3\text{K}_2\text{SO}_4 + 2\text{Fe(OH)}_3}
\]
---
12. \( \text{FeS}_2 + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 + \text{SO}_2 \)
- Start with the unbalanced equation:
\[
\text{FeS}_2 + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 + \text{SO}_2
\]
- On the left side: 1 Fe, 2 S, and 2 O.
- On the right side: 2 Fe, 3 O in Fe₂O₃, and 1 S and 2 O in SO₂.
- To balance iron (Fe), we need 2 FeS₂ molecules on the left:
\[
2\text{FeS}_2 + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 + \text{SO}_2
\]
- Now, there are 4 S atoms on the left side, so we need 4 SO₂ molecules on the right:
\[
2\text{FeS}_2 + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 + 4\text{SO}_2
\]
- Now, there are 8 O atoms on the right side, so we need 11/2 O₂ molecules on the left:
\[
2\text{FeS}_2 + \frac{11}{2}\text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 + 4\text{SO}_2
\]
Multiply through by 2 to eliminate the fraction:
\[
4\text{FeS}_2 + 11\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2
\]
- Final balanced equation:
\[
\boxed{4\text{FeS}_2 + 11\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2}
\]
---
13. \( \text{Al} + \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe} \)
- Start with the unbalanced equation:
\[
\text{Al} + \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}
\]
- On the left side: 1 Al, 1 Fe, and 1 O.
- On the right side: 2 Al, 3 O in Al₂O₃, and 1 Fe.
- To balance aluminum (Al), we need 2 Al atoms on the left:
\[
2\text{Al} + \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}
\]
- Now, there are 3 O atoms on the right side, so we need 3 FeO molecules on the left:
\[
2\text{Al} + 3\text{FeO} \rightarrow \text{Al}_2\text{O}_3 + 3\text{Fe}
\]
- Final balanced equation:
\[
\boxed{2\text{Al} + 3\text{FeO} \rightarrow \text{Al}_2\text{O}_3 + 3\text{Fe}}
\]
---
14. \( \text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \)
- Start with the unbalanced equation:
\[
\text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2
\]
- On the left side: 2 Na, 1 C, 3 O, 1 H, and 1 Cl.
- On the right side: 1 Na, 1 Cl in NaCl, 2 H and 1 O in H₂O, and 1 C and 2 O in CO₂.
- To balance sodium (Na), we need 2 NaCl molecules on the right:
\[
\text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2
\]
- Now, there are 2 Cl atoms on the right side, so we need 2 HCl molecules on the left:
\[
\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2
\]
- Final balanced equation:
\[
\boxed{\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2}
\]
---
15. \( \text{K} + \text{Br}_2 \rightarrow \text{KBr} \)
- Start with the unbalanced equation:
\[
\text{K} + \text{Br}_2 \rightarrow \text{KBr}
\]
- On the left side: 1 K and 2 Br.
- On the right side: 1 K and 1 Br in KBr.
- To balance bromine (Br), we need 2 KBr molecules on the right:
\[
\text{K} + \text{Br}_2 \rightarrow 2\text{KBr}
\]
- Now, there are 2 K atoms on the right side, so we need 2 K atoms on the left:
\[
2\text{K} + \text{Br}_2 \rightarrow 2\text{KBr}
\]
- Final balanced equation:
\[
\boxed{2\text{K} + \text{Br}_2 \rightarrow 2\text{KBr}}
\]
---
16. \( \text{P}_4 + \text{O}_2 \rightarrow \text{P}_2\text{O}_5 \)
- Start with the unbalanced equation:
\[
\text{P}_4 + \text{O}_2 \rightarrow \text{P}_2\text{O}_5
\]
- On the left side: 4 P and 2 O.
- On the right side: 2 P and 5 O in P₂O₅.
- To balance phosphorus (P), we need 2 P₂O₅ molecules on the right:
\[
\text{P}_4 + \text{O}_2 \rightarrow 2\text{P}_2\text{O}_5
\]
- Now, there are 10 O atoms on the right side, so we need 5 O₂ molecules on the left:
\[
\text{P}_4 + 5\text{O}_2 \rightarrow 2\text{P}_2\text{O}_5
\]
- Final balanced equation:
\[
\boxed{\text{P}_4 + 5\text{O}_2 \rightarrow 2\text{P}_2\text{O}_5}
\]
---
17. \( \text{C}_2\text{H}_2 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \)
- Start with the unbalanced equation:
\[
\text{C}_2\text{H}_2 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
\]
- On the left side: 2 C, 2 H, and 2 O.
- On the right side: 1 C, 2 O in CO₂, and 2 H and 1 O in H₂O.
- To balance carbon (C), we need 2 CO₂ molecules on the right:
\[
\text{C}_2\text{H}_2 + \text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O}
\]
- Now, there are 4 O atoms on the right side from CO₂ and 1 O atom from H₂O, so we need 5/2 O₂ molecules on the left:
\[
\text{C}_2\text{H}_2 + \frac{5}{2}\text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O}
\]
Multiply through by 2 to eliminate the fraction:
\[
2\text{C}_2\text{H}_2 + 5\text{O}_2 \rightarrow 4\text{CO}_2 + 2\text{H}_2\text{O}
\]
- Final balanced equation:
\[
\boxed{2\text{C}_2\text{H}_2 + 5\text{O}_2 \rightarrow 4\text{CO}_2 + 2\text{H}_2\text{O}}
\]
---
Final Answer:
\[
\boxed{
\begin{aligned}
1. & \quad \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \\
2. & \quad \text{S}_8 + 12\text{O}_2 \rightarrow 8\text{SO}_3 \\
3. & \quad 2\text{HgO} \rightarrow 2\text{Hg} + \text{O}_2 \\
4. & \quad \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \\
5. & \quad \text{SiCl}_4 + 4\text{H}_2\text{O} \rightarrow \text{H}_4\text{SiO}_4 + 4\text{HCl} \\
6. & \quad 2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 \\
7. & \quad 2\text{H}_3\text{PO}_4 \rightarrow \text{H}_4\text{P}_2\text{O}_7 + \text{H}_2\text{O} \\
8. & \quad 2\text{Si}_2\text{H}_3 + 7\text{O}_2 \rightarrow 4\text{SiO}_2 + 6\text{H}_2\text{O} \\
9. & \quad 2\text{Al(OH)}_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6\text{H}_2\text{O} \\
10. & \quad 4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 \\
11. & \quad \text{Fe}_2(\text{SO}_4)_3 + 6\text{KOH} \rightarrow 3\text{K}_2\text{SO}_4 + 2\text{Fe(OH)}_3 \\
12. & \quad 4\text{FeS}_2 + 11\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2 \\
13. & \quad 2\text{Al} + 3\text{FeO} \rightarrow \text{Al}_2\text{O}_3 + 3\text{Fe} \\
14. & \quad \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \\
15. & \quad 2\text{K} + \text{Br}_2 \rightarrow 2\text{KBr} \\
16. & \quad \text{P}_4 + 5\text{O}_2 \rightarrow 2\text{P}_2\text{O}_5 \\
17. & \quad 2\text{C}_2\text{H}_2 + 5\text{O}_2 \rightarrow 4\text{CO}_2 + 2\text{H}_2\text{O}
\end{aligned}
}
\]
Parent Tip: Review the logic above to help your child master the concept of chemistry worksheet balancing equations.
![49 Balancing Chemical Equations Worksheets [with Answers]](/images2/chemistry-worksheet-balancing-equations/chemistry-worksheet-balancing-equations-7.jpg)
