49 Balancing Chemical Equations Worksheets [with Answers] - Free Printable
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Step-by-step solution for: 49 Balancing Chemical Equations Worksheets [with Answers]
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Show Answer Key & Explanations
Step-by-step solution for: 49 Balancing Chemical Equations Worksheets [with Answers]
To balance 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:
\[
\text{N}_2 + \text{H}_2 \rightarrow \text{NH}_3
\]
- There are 2 nitrogen (N) atoms on the left and 1 N atom on the right. To balance N, multiply \(\text{NH}_3\) by 2:
\[
\text{N}_2 + \text{H}_2 \rightarrow 2\text{NH}_3
\]
- Now there are 6 hydrogen (H) atoms on the right and 2 H atoms on the left. To balance H, multiply \(\text{H}_2\) by 3:
\[
\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3
\]
- Final balanced equation:
\[
\boxed{\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3}
\]
---
- Start with the unbalanced equation:
\[
\text{S}_8 + \text{O}_2 \rightarrow \text{SO}_3
\]
- There are 8 sulfur (S) atoms on the left and 1 S atom on the right. Multiply \(\text{SO}_3\) by 8:
\[
\text{S}_8 + \text{O}_2 \rightarrow 8\text{SO}_3
\]
- Now there are 24 oxygen (O) atoms on the right and 2 O atoms on the left. Multiply \(\text{O}_2\) by 12:
\[
\text{S}_8 + 12\text{O}_2 \rightarrow 8\text{SO}_3
\]
- Final balanced equation:
\[
\boxed{\text{S}_8 + 12\text{O}_2 \rightarrow 8\text{SO}_3}
\]
---
- Start with the unbalanced equation:
\[
\text{HgO} \rightarrow \text{Hg} + \text{O}_2
\]
- There is 1 oxygen (O) atom on the left and 2 O atoms on the right. Multiply \(\text{HgO}\) by 2:
\[
2\text{HgO} \rightarrow 2\text{Hg} + \text{O}_2
\]
- Final balanced equation:
\[
\boxed{2\text{HgO} \rightarrow 2\text{Hg} + \text{O}_2}
\]
---
- Start with the unbalanced equation:
\[
\text{Zn} + \text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2
\]
- There are 2 chlorine (Cl) atoms on the right and 1 Cl atom on the left. Multiply \(\text{HCl}\) by 2:
\[
\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2
\]
- Final balanced equation:
\[
\boxed{\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2}
\]
---
- Start with the unbalanced equation:
\[
\text{SiCl}_4 + \text{H}_2\text{O} \rightarrow \text{H}_4\text{SiO}_4 + \text{HCl}
\]
- There are 4 chlorine (Cl) atoms on the left and 1 Cl atom on the right. Multiply \(\text{HCl}\) by 4:
\[
\text{SiCl}_4 + \text{H}_2\text{O} \rightarrow \text{H}_4\text{SiO}_4 + 4\text{HCl}
\]
- Now there are 4 hydrogen (H) atoms on the left and 8 H atoms on the right. Multiply \(\text{H}_2\text{O}\) by 4:
\[
\text{SiCl}_4 + 4\text{H}_2\text{O} \rightarrow \text{H}_4\text{SiO}_4 + 4\text{HCl}
\]
- 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:
\[
\text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2
\]
- There are 2 hydrogen (H) atoms on the right and 2 H atoms on the left. Multiply \(\text{Na}\) and \(\text{NaOH}\) by 2:
\[
2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2
\]
- 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:
\[
\text{H}_3\text{PO}_4 \rightarrow \text{H}_4\text{P}_2\text{O}_7 + \text{H}_2\text{O}
\]
- There are 2 phosphorus (P) atoms on the right and 1 P atom on the left. Multiply \(\text{H}_3\text{PO}_4\) by 2:
\[
2\text{H}_3\text{PO}_4 \rightarrow \text{H}_4\text{P}_2\text{O}_7 + \text{H}_2\text{O}
\]
- Now there are 6 hydrogen (H) atoms on the left and 6 H atoms on the right. Multiply \(\text{H}_2\text{O}\) by 1:
\[
2\text{H}_3\text{PO}_4 \rightarrow \text{H}_4\text{P}_2\text{O}_7 + \text{H}_2\text{O}
\]
- 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:
\[
\text{Si}_2\text{H}_3 + \text{O}_2 \rightarrow \text{SiO}_2 + \text{H}_2\text{O}
\]
- There are 2 silicon (Si) atoms on the left and 1 Si atom on the right. Multiply \(\text{SiO}_2\) by 2:
\[
\text{Si}_2\text{H}_3 + \text{O}_2 \rightarrow 2\text{SiO}_2 + \text{H}_2\text{O}
\]
- Now there are 3 hydrogen (H) atoms on the left and 2 H atoms on the right. Multiply \(\text{H}_2\text{O}\) by 3/2 (or 3 and adjust later):
\[
\text{Si}_2\text{H}_3 + \text{O}_2 \rightarrow 2\text{SiO}_2 + \frac{3}{2}\text{H}_2\text{O}
\]
- Multiply everything by 2 to clear the fraction:
\[
2\text{Si}_2\text{H}_3 + 2\text{O}_2 \rightarrow 4\text{SiO}_2 + 3\text{H}_2\text{O}
\]
- 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:
\[
\text{Al(OH)}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O}
\]
- There are 2 aluminum (Al) atoms on the right and 1 Al atom on the left. Multiply \(\text{Al(OH)}_3\) by 2:
\[
2\text{Al(OH)}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O}
\]
- Now there are 6 hydroxide (OH) groups on the left and 0 OH groups on the right. Multiply \(\text{H}_2\text{O}\) by 6:
\[
2\text{Al(OH)}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6\text{H}_2\text{O}
\]
- Balance sulfur (S) and oxygen (O):
\[
2\text{Al(OH)}_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6\text{H}_2\text{O}
\]
- 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:
\[
\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3
\]
- There are 2 iron (Fe) atoms on the right and 1 Fe atom on the left. Multiply \(\text{Fe}\) by 2:
\[
2\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3
\]
- Now there are 3 oxygen (O) atoms on the right and 2 O atoms on the left. Multiply \(\text{O}_2\) by 3/2 (or 3 and adjust later):
\[
2\text{Fe} + \frac{3}{2}\text{O}_2 \rightarrow \text{Fe}_2\text{O}_3
\]
- Multiply everything by 2 to clear 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:
\[
\text{Fe}_2(\text{SO}_4)_3 + \text{KOH} \rightarrow \text{K}_2\text{SO}_4 + \text{Fe(OH)}_3
\]
- There are 2 iron (Fe) atoms on the left and 1 Fe atom on the right. Multiply \(\text{Fe(OH)}_3\) by 2:
\[
\text{Fe}_2(\text{SO}_4)_3 + \text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{Fe(OH)}_3
\]
- Now there are 3 sulfate (\(\text{SO}_4\)) groups on the left and 1 \(\text{SO}_4\) group on the right. Multiply \(\text{K}_2\text{SO}_4\) by 3:
\[
\text{Fe}_2(\text{SO}_4)_3 + \text{KOH} \rightarrow 3\text{K}_2\text{SO}_4 + 2\text{Fe(OH)}_3
\]
- Balance potassium (K) and hydroxide (OH):
\[
\text{Fe}_2(\text{SO}_4)_3 + 6\text{KOH} \rightarrow 3\text{K}_2\text{SO}_4 + 2\text{Fe(OH)}_3
\]
- 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:
\[
\text{FeS}_2 + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 + \text{SO}_2
\]
- There are 2 iron (Fe) atoms on the left and 2 Fe atoms on the right. Multiply \(\text{FeS}_2\) by 4:
\[
4\text{FeS}_2 + \text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + \text{SO}_2
\]
- Now there are 8 sulfur (S) atoms on the left and 1 S atom on the right. Multiply \(\text{SO}_2\) by 8:
\[
4\text{FeS}_2 + \text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2
\]
- Balance oxygen (O):
\[
4\text{FeS}_2 + 11\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2
\]
- 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:
\[
\text{Al} + \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}
\]
- There are 2 aluminum (Al) atoms on the right and 1 Al atom on the left. Multiply \(\text{Al}\) by 2:
\[
2\text{Al} + \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}
\]
- Now there are 1 iron (Fe) atom on the left and 1 Fe atom on the right. Multiply \(\text{FeO}\) by 3:
\[
2\text{Al} + 3\text{FeO} \rightarrow \text{Al}_2\text{O}_3 + 3\text{Fe}
\]
- 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:
\[
\text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2
\]
- There are 2 sodium (Na) atoms on the left and 1 Na atom on the right. Multiply \(\text{NaCl}\) by 2:
\[
\text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2
\]
- Now there are 2 chloride (Cl) atoms on the right and 1 Cl atom on the left. Multiply \(\text{HCl}\) by 2:
\[
\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2
\]
- 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:
\[
\text{K} + \text{Br}_2 \rightarrow \text{KBr}
\]
- There are 2 bromine (Br) atoms on the left and 1 Br atom on the right. Multiply \(\text{KBr}\) by 2:
\[
\text{K} + \text{Br}_2 \rightarrow 2\text{KBr}
\]
- Now there are 2 potassium (K) atoms on the right and 1 K atom on the left. Multiply \(\text{K}\) by 2:
\[
2\text{K} + \text{Br}_2 \rightarrow 2\text{KBr}
\]
- Final balanced equation:
\[
\boxed{2\text{K} + \text{Br}_2 \rightarrow 2\text{KBr}}
\]
---
- Start with the unbalanced equation:
\[
\text{P}_4 + \text{O}_2 \rightarrow \text{P}_2\text{O}_5
\]
- There are 4 phosphorus (P) atoms on the left and 2 P atoms on the right. Multiply \(\text{P}_2\text{O}_5\) by 2:
\[
\text{P}_4 + \text{O}_2 \rightarrow 2\text{P}_2\text{O}_5
\]
- Now there are 10 oxygen (O) atoms on the right and 2 O atoms on the left. Multiply \(\text{O}_2\) by 5:
\[
\text{P}_4 + 5\text{O}_2 \rightarrow 2\text{P}_2\text{O}_5
\]
- Final balanced equation:
\[
\boxed{\text{P}_4 + 5\text{O}_2 \rightarrow 2\text{P}_2\text{O}_5}
\]
---
- Start with the unbalanced equation:
\[
\text{C}_2\text{H}_2 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
\]
- There are 2 carbon (C) atoms on the left and 1 C atom on the right. Multiply \(\text{CO}_2\) by 2:
\[
\text{C}_2\text{H}_2 + \text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O}
\]
- Now there are 2 hydrogen (H) atoms on the left and 2 H atoms on the right. Multiply \(\text{H}_2\text{O}\) by 1:
\[
\text{C}_2\text{H}_2 + \text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O}
\]
- Balance oxygen (O):
\[
\text{C}_2\text{H}_2 + \frac{5}{2}\text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O}
\]
- Multiply everything by 2 to clear 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:
\[
\text{N}_2 + \text{H}_2 \rightarrow \text{NH}_3
\]
- There are 2 nitrogen (N) atoms on the left and 1 N atom on the right. To balance N, multiply \(\text{NH}_3\) by 2:
\[
\text{N}_2 + \text{H}_2 \rightarrow 2\text{NH}_3
\]
- Now there are 6 hydrogen (H) atoms on the right and 2 H atoms on the left. To balance H, multiply \(\text{H}_2\) by 3:
\[
\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3
\]
- 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:
\[
\text{S}_8 + \text{O}_2 \rightarrow \text{SO}_3
\]
- There are 8 sulfur (S) atoms on the left and 1 S atom on the right. Multiply \(\text{SO}_3\) by 8:
\[
\text{S}_8 + \text{O}_2 \rightarrow 8\text{SO}_3
\]
- Now there are 24 oxygen (O) atoms on the right and 2 O atoms on the left. Multiply \(\text{O}_2\) by 12:
\[
\text{S}_8 + 12\text{O}_2 \rightarrow 8\text{SO}_3
\]
- 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:
\[
\text{HgO} \rightarrow \text{Hg} + \text{O}_2
\]
- There is 1 oxygen (O) atom on the left and 2 O atoms on the right. Multiply \(\text{HgO}\) by 2:
\[
2\text{HgO} \rightarrow 2\text{Hg} + \text{O}_2
\]
- 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:
\[
\text{Zn} + \text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2
\]
- There are 2 chlorine (Cl) atoms on the right and 1 Cl atom on the left. Multiply \(\text{HCl}\) by 2:
\[
\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2
\]
- 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:
\[
\text{SiCl}_4 + \text{H}_2\text{O} \rightarrow \text{H}_4\text{SiO}_4 + \text{HCl}
\]
- There are 4 chlorine (Cl) atoms on the left and 1 Cl atom on the right. Multiply \(\text{HCl}\) by 4:
\[
\text{SiCl}_4 + \text{H}_2\text{O} \rightarrow \text{H}_4\text{SiO}_4 + 4\text{HCl}
\]
- Now there are 4 hydrogen (H) atoms on the left and 8 H atoms on the right. Multiply \(\text{H}_2\text{O}\) by 4:
\[
\text{SiCl}_4 + 4\text{H}_2\text{O} \rightarrow \text{H}_4\text{SiO}_4 + 4\text{HCl}
\]
- 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:
\[
\text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2
\]
- There are 2 hydrogen (H) atoms on the right and 2 H atoms on the left. Multiply \(\text{Na}\) and \(\text{NaOH}\) by 2:
\[
2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2
\]
- 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:
\[
\text{H}_3\text{PO}_4 \rightarrow \text{H}_4\text{P}_2\text{O}_7 + \text{H}_2\text{O}
\]
- There are 2 phosphorus (P) atoms on the right and 1 P atom on the left. Multiply \(\text{H}_3\text{PO}_4\) by 2:
\[
2\text{H}_3\text{PO}_4 \rightarrow \text{H}_4\text{P}_2\text{O}_7 + \text{H}_2\text{O}
\]
- Now there are 6 hydrogen (H) atoms on the left and 6 H atoms on the right. Multiply \(\text{H}_2\text{O}\) by 1:
\[
2\text{H}_3\text{PO}_4 \rightarrow \text{H}_4\text{P}_2\text{O}_7 + \text{H}_2\text{O}
\]
- 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:
\[
\text{Si}_2\text{H}_3 + \text{O}_2 \rightarrow \text{SiO}_2 + \text{H}_2\text{O}
\]
- There are 2 silicon (Si) atoms on the left and 1 Si atom on the right. Multiply \(\text{SiO}_2\) by 2:
\[
\text{Si}_2\text{H}_3 + \text{O}_2 \rightarrow 2\text{SiO}_2 + \text{H}_2\text{O}
\]
- Now there are 3 hydrogen (H) atoms on the left and 2 H atoms on the right. Multiply \(\text{H}_2\text{O}\) by 3/2 (or 3 and adjust later):
\[
\text{Si}_2\text{H}_3 + \text{O}_2 \rightarrow 2\text{SiO}_2 + \frac{3}{2}\text{H}_2\text{O}
\]
- Multiply everything by 2 to clear the fraction:
\[
2\text{Si}_2\text{H}_3 + 2\text{O}_2 \rightarrow 4\text{SiO}_2 + 3\text{H}_2\text{O}
\]
- 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:
\[
\text{Al(OH)}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O}
\]
- There are 2 aluminum (Al) atoms on the right and 1 Al atom on the left. Multiply \(\text{Al(OH)}_3\) by 2:
\[
2\text{Al(OH)}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O}
\]
- Now there are 6 hydroxide (OH) groups on the left and 0 OH groups on the right. Multiply \(\text{H}_2\text{O}\) by 6:
\[
2\text{Al(OH)}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6\text{H}_2\text{O}
\]
- Balance sulfur (S) and oxygen (O):
\[
2\text{Al(OH)}_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6\text{H}_2\text{O}
\]
- 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:
\[
\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3
\]
- There are 2 iron (Fe) atoms on the right and 1 Fe atom on the left. Multiply \(\text{Fe}\) by 2:
\[
2\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3
\]
- Now there are 3 oxygen (O) atoms on the right and 2 O atoms on the left. Multiply \(\text{O}_2\) by 3/2 (or 3 and adjust later):
\[
2\text{Fe} + \frac{3}{2}\text{O}_2 \rightarrow \text{Fe}_2\text{O}_3
\]
- Multiply everything by 2 to clear 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:
\[
\text{Fe}_2(\text{SO}_4)_3 + \text{KOH} \rightarrow \text{K}_2\text{SO}_4 + \text{Fe(OH)}_3
\]
- There are 2 iron (Fe) atoms on the left and 1 Fe atom on the right. Multiply \(\text{Fe(OH)}_3\) by 2:
\[
\text{Fe}_2(\text{SO}_4)_3 + \text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{Fe(OH)}_3
\]
- Now there are 3 sulfate (\(\text{SO}_4\)) groups on the left and 1 \(\text{SO}_4\) group on the right. Multiply \(\text{K}_2\text{SO}_4\) by 3:
\[
\text{Fe}_2(\text{SO}_4)_3 + \text{KOH} \rightarrow 3\text{K}_2\text{SO}_4 + 2\text{Fe(OH)}_3
\]
- Balance potassium (K) and hydroxide (OH):
\[
\text{Fe}_2(\text{SO}_4)_3 + 6\text{KOH} \rightarrow 3\text{K}_2\text{SO}_4 + 2\text{Fe(OH)}_3
\]
- 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:
\[
\text{FeS}_2 + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 + \text{SO}_2
\]
- There are 2 iron (Fe) atoms on the left and 2 Fe atoms on the right. Multiply \(\text{FeS}_2\) by 4:
\[
4\text{FeS}_2 + \text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + \text{SO}_2
\]
- Now there are 8 sulfur (S) atoms on the left and 1 S atom on the right. Multiply \(\text{SO}_2\) by 8:
\[
4\text{FeS}_2 + \text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2
\]
- Balance oxygen (O):
\[
4\text{FeS}_2 + 11\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2
\]
- 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:
\[
\text{Al} + \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}
\]
- There are 2 aluminum (Al) atoms on the right and 1 Al atom on the left. Multiply \(\text{Al}\) by 2:
\[
2\text{Al} + \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}
\]
- Now there are 1 iron (Fe) atom on the left and 1 Fe atom on the right. Multiply \(\text{FeO}\) by 3:
\[
2\text{Al} + 3\text{FeO} \rightarrow \text{Al}_2\text{O}_3 + 3\text{Fe}
\]
- 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:
\[
\text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2
\]
- There are 2 sodium (Na) atoms on the left and 1 Na atom on the right. Multiply \(\text{NaCl}\) by 2:
\[
\text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2
\]
- Now there are 2 chloride (Cl) atoms on the right and 1 Cl atom on the left. Multiply \(\text{HCl}\) by 2:
\[
\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2
\]
- 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:
\[
\text{K} + \text{Br}_2 \rightarrow \text{KBr}
\]
- There are 2 bromine (Br) atoms on the left and 1 Br atom on the right. Multiply \(\text{KBr}\) by 2:
\[
\text{K} + \text{Br}_2 \rightarrow 2\text{KBr}
\]
- Now there are 2 potassium (K) atoms on the right and 1 K atom on the left. Multiply \(\text{K}\) by 2:
\[
2\text{K} + \text{Br}_2 \rightarrow 2\text{KBr}
\]
- 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:
\[
\text{P}_4 + \text{O}_2 \rightarrow \text{P}_2\text{O}_5
\]
- There are 4 phosphorus (P) atoms on the left and 2 P atoms on the right. Multiply \(\text{P}_2\text{O}_5\) by 2:
\[
\text{P}_4 + \text{O}_2 \rightarrow 2\text{P}_2\text{O}_5
\]
- Now there are 10 oxygen (O) atoms on the right and 2 O atoms on the left. Multiply \(\text{O}_2\) by 5:
\[
\text{P}_4 + 5\text{O}_2 \rightarrow 2\text{P}_2\text{O}_5
\]
- 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:
\[
\text{C}_2\text{H}_2 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
\]
- There are 2 carbon (C) atoms on the left and 1 C atom on the right. Multiply \(\text{CO}_2\) by 2:
\[
\text{C}_2\text{H}_2 + \text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O}
\]
- Now there are 2 hydrogen (H) atoms on the left and 2 H atoms on the right. Multiply \(\text{H}_2\text{O}\) by 1:
\[
\text{C}_2\text{H}_2 + \text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O}
\]
- Balance oxygen (O):
\[
\text{C}_2\text{H}_2 + \frac{5}{2}\text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O}
\]
- Multiply everything by 2 to clear 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 chemical balancing worksheet with answers.
![49 Balancing Chemical Equations Worksheets [with Answers]](/images2/chemical-balancing-worksheet-with-answers/chemical-balancing-worksheet-with-answers-0.jpg)
