Balancing chemical equations worksheet for practice in stoichiometry.
A worksheet with ten chemical equations to balance, including reactions involving iron, phosphorus, aluminum, and other elements, with blank spaces for coefficients.
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Step-by-step solution for: Balancing Chemical Equations Worksheet | STEM Sheets
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Show Answer Key & Explanations
Step-by-step solution for: Balancing Chemical Equations Worksheet | STEM Sheets
To solve the given chemical equations, we need to balance them by ensuring that the number of atoms of each element is the same on both sides of the equation. Let's go through each equation step by step.
---
- Reactants: Fe, O₂
- Products: Fe₂O₃
#### Balancing:
- Start with Fe: There are 2 Fe atoms in Fe₂O₃, so we need 2 Fe atoms on the left.
- Next, O: There are 3 O atoms in Fe₂O₃, and O₂ provides 2 O atoms. To get 3 O atoms, we need 1.5 O₂ molecules, but since we can't have fractional coefficients, multiply everything by 2 to clear the fraction.
Final balanced equation:
$$
4 \text{Fe} + 3 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3
$$
---
- Reactants: P₄O₆
- Products: P₄, P₂O₄
#### Balancing:
- P: There are 4 P atoms in P₄O₆. On the right, there are 4 P atoms in P₄ and 2 P atoms in P₂O₄. To balance P, we need 2 P₂O₄ molecules.
- O: There are 6 O atoms in P₄O₆. On the right, there are 4 O atoms in 2 P₂O₄ molecules. This is already balanced.
Final balanced equation:
$$
P_4O_6 \rightarrow P_4 + 2P_2O_4
$$
---
- Reactants: Fe(OH)₃
- Products: Fe₂O₃, H₂O
#### Balancing:
- Fe: There are 2 Fe atoms in Fe₂O₃, so we need 2 Fe(OH)₃ molecules.
- O: There are 9 O atoms in 2 Fe(OH)₃ molecules (3 O atoms per molecule). On the right, there are 3 O atoms in Fe₂O₃ and 2 O atoms in 3 H₂O molecules. This is balanced.
- H: There are 6 H atoms in 2 Fe(OH)₃ molecules. On the right, there are 6 H atoms in 3 H₂O molecules. This is balanced.
Final balanced equation:
$$
2 \text{Fe(OH)}_3 \rightarrow \text{Fe}_2\text{O}_3 + 3 \text{H}_2\text{O}
$$
---
- Reactants: UO₃, H₂
- Products: UO₂, H₂O
#### Balancing:
- U: There is 1 U atom on both sides, so it is balanced.
- O: There are 3 O atoms in UO₃. On the right, there are 2 O atoms in UO₂ and 1 O atom in H₂O. To balance O, we need 2 H₂O molecules.
- H: There are 4 H atoms in 2 H₂ molecules. On the right, there are 4 H atoms in 2 H₂O molecules. This is balanced.
Final balanced equation:
$$
UO_3 + 2H_2 \rightarrow UO_2 + 2H_2O
$$
---
- Reactants: V₂O₅, Al
- Products: Al₂O₃, V
#### Balancing:
- V: There are 2 V atoms in V₂O₅, so we need 2 V atoms on the right.
- O: There are 5 O atoms in V₂O₅. On the right, there are 3 O atoms in Al₂O₃. To balance O, we need 5/3 Al₂O₃ molecules, but since we can't have fractional coefficients, multiply everything by 3.
- Al: There are 2 Al atoms in Al₂O₃, so we need 6 Al atoms on the left.
Final balanced equation:
$$
3V_2O_5 + 10Al \rightarrow 5Al_2O_3 + 6V
$$
---
- Reactants: PCl₅, KNO₂
- Products: NOCl, POCl₃, KCl
#### Balancing:
- P: There is 1 P atom in PCl₅, so we need 1 P atom in POCl₃.
- Cl: There are 5 Cl atoms in PCl₅. On the right, there is 1 Cl atom in NOCl, 3 Cl atoms in POCl₃, and 1 Cl atom in KCl. This is balanced.
- K: There is 1 K atom in KNO₂, so we need 1 K atom in KCl.
- N: There is 1 N atom in KNO₂, so we need 1 N atom in NOCl.
- O: There are 2 O atoms in KNO₂. On the right, there is 1 O atom in NOCl and 3 O atoms in POCl₃. This is balanced.
Final balanced equation:
$$
PCl_5 + KNO_2 \rightarrow NOCl + POCl_3 + KCl
$$
---
- Reactants: NaOH, Zn(NO₃)₂
- Products: NaNO₃, Zn(OH)₂
#### Balancing:
- Na: There is 1 Na atom in NaOH and 1 Na atom in NaNO₃, so we need 2 NaOH molecules.
- Zn: There is 1 Zn atom in Zn(NO₃)₂ and 1 Zn atom in Zn(OH)₂, so it is balanced.
- N: There are 2 N atoms in Zn(NO₃)₂ and 2 N atoms in 2 NaNO₃, so it is balanced.
- O: There are 6 O atoms in Zn(NO₃)₂ and 2 O atoms in 2 NaOH. On the right, there are 3 O atoms in 2 NaNO₃ and 2 O atoms in Zn(OH)₂. This is balanced.
- H: There are 2 H atoms in 2 NaOH and 2 H atoms in Zn(OH)₂. This is balanced.
Final balanced equation:
$$
2NaOH + Zn(NO_3)_2 \rightarrow 2NaNO_3 + Zn(OH)_2
$$
---
- Reactants: Bi(NO₃)₃, H₂S
- Products: Bi₂S₃, HNO₃
#### Balancing:
- Bi: There are 2 Bi atoms in Bi₂S₃, so we need 2 Bi(NO₃)₃ molecules.
- S: There are 3 S atoms in Bi₂S₃, so we need 3 H₂S molecules.
- N: There are 6 N atoms in 2 Bi(NO₃)₃, so we need 6 HNO₃ molecules.
- O: There are 18 O atoms in 2 Bi(NO₃)₃. On the right, there are 3 O atoms in Bi₂S₃ and 3 O atoms in each HNO₃ molecule. This is balanced.
- H: There are 6 H atoms in 3 H₂S molecules. On the right, there are 6 H atoms in 6 HNO₃ molecules. This is balanced.
Final balanced equation:
$$
2Bi(NO_3)_3 + 3H_2S \rightarrow Bi_2S_3 + 6HNO_3
$$
---
- Reactants: MgNH₄PO₄
- Products: Mg₂P₂O₇, NH₃, H₂O
#### Balancing:
- Mg: There are 2 Mg atoms in Mg₂P₂O₇, so we need 2 MgNH₄PO₄ molecules.
- P: There are 2 P atoms in Mg₂P₂O₇, so we need 2 MgNH₄PO₄ molecules.
- O: There are 7 O atoms in Mg₂P₂O₇. On the right, there are 4 O atoms in 2 H₂O molecules. This is balanced.
- N: There is 1 N atom in NH₃, so we need 2 NH₃ molecules.
- H: There are 8 H atoms in 2 MgNH₄PO₄ molecules. On the right, there are 6 H atoms in 2 NH₃ molecules and 4 H atoms in 2 H₂O molecules. This is balanced.
Final balanced equation:
$$
2MgNH_4PO_4 \rightarrow Mg_2P_2O_7 + 2NH_3 + 2H_2O
$$
---
- Reactants: Al, NH₄ClO₄
- Products: Al₂O₃, AlCl₃, NO, H₂O
#### Balancing:
- Al: There are 2 Al atoms in Al₂O₃ and 3 Al atoms in AlCl₃, so we need 5 Al atoms on the left.
- Cl: There are 4 Cl atoms in NH₄ClO₄, so we need 4 AlCl₃ molecules.
- O: There are 4 O atoms in NH₄ClO₄. On the right, there are 3 O atoms in Al₂O₃, 1 O atom in NO, and 1 O atom in H₂O. To balance O, we need 6 H₂O molecules.
- N: There is 1 N atom in NH₄ClO₄, so we need 1 NO molecule.
- H: There are 4 H atoms in NH₄ClO₄. On the right, there are 2 H atoms in 2 H₂O molecules. To balance H, we need 2 NH₄ClO₄ molecules.
Final balanced equation:
$$
10Al + 6NH_4ClO_4 \rightarrow 2Al_2O_3 + 4AlCl_3 + 6NO + 12H_2O
$$
---
1. $ 4 \text{Fe} + 3 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 $
2. $ P_4O_6 \rightarrow P_4 + 2P_2O_4 $
3. $ 2 \text{Fe(OH)}_3 \rightarrow \text{Fe}_2\text{O}_3 + 3 \text{H}_2\text{O} $
4. $ UO_3 + 2H_2 \rightarrow UO_2 + 2H_2O $
5. $ 3V_2O_5 + 10Al \rightarrow 5Al_2O_3 + 6V $
6. $ PCl_5 + KNO_2 \rightarrow NOCl + POCl_3 + KCl $
7. $ 2NaOH + Zn(NO_3)_2 \rightarrow 2NaNO_3 + Zn(OH)_2 $
8. $ 2Bi(NO_3)_3 + 3H_2S \rightarrow Bi_2S_3 + 6HNO_3 $
9. $ 2MgNH_4PO_4 \rightarrow Mg_2P_2O_7 + 2NH_3 + 2H_2O $
10. $ 10Al + 6NH_4ClO_4 \rightarrow 2Al_2O_3 + 4AlCl_3 + 6NO + 12H_2O $
$$
\boxed{
\begin{aligned}
&1. \ 4 \text{Fe} + 3 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 \\
&2. \ P_4O_6 \rightarrow P_4 + 2P_2O_4 \\
&3. \ 2 \text{Fe(OH)}_3 \rightarrow \text{Fe}_2\text{O}_3 + 3 \text{H}_2\text{O} \\
&4. \ UO_3 + 2H_2 \rightarrow UO_2 + 2H_2O \\
&5. \ 3V_2O_5 + 10Al \rightarrow 5Al_2O_3 + 6V \\
&6. \ PCl_5 + KNO_2 \rightarrow NOCl + POCl_3 + KCl \\
&7. \ 2NaOH + Zn(NO_3)_2 \rightarrow 2NaNO_3 + Zn(OH)_2 \\
&8. \ 2Bi(NO_3)_3 + 3H_2S \rightarrow Bi_2S_3 + 6HNO_3 \\
&9. \ 2MgNH_4PO_4 \rightarrow Mg_2P_2O_7 + 2NH_3 + 2H_2O \\
&10. \ 10Al + 6NH_4ClO_4 \rightarrow 2Al_2O_3 + 4AlCl_3 + 6NO + 12H_2O \\
\end{aligned}
}
$$
---
1) Fe + O₂ → Fe₂O₃
- Reactants: Fe, O₂
- Products: Fe₂O₃
#### Balancing:
- Start with Fe: There are 2 Fe atoms in Fe₂O₃, so we need 2 Fe atoms on the left.
- Next, O: There are 3 O atoms in Fe₂O₃, and O₂ provides 2 O atoms. To get 3 O atoms, we need 1.5 O₂ molecules, but since we can't have fractional coefficients, multiply everything by 2 to clear the fraction.
Final balanced equation:
$$
4 \text{Fe} + 3 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3
$$
---
2) P₄O₆ → P₄ + P₂O₄
- Reactants: P₄O₆
- Products: P₄, P₂O₄
#### Balancing:
- P: There are 4 P atoms in P₄O₆. On the right, there are 4 P atoms in P₄ and 2 P atoms in P₂O₄. To balance P, we need 2 P₂O₄ molecules.
- O: There are 6 O atoms in P₄O₆. On the right, there are 4 O atoms in 2 P₂O₄ molecules. This is already balanced.
Final balanced equation:
$$
P_4O_6 \rightarrow P_4 + 2P_2O_4
$$
---
3) Fe(OH)₃ → Fe₂O₃ + H₂O
- Reactants: Fe(OH)₃
- Products: Fe₂O₃, H₂O
#### Balancing:
- Fe: There are 2 Fe atoms in Fe₂O₃, so we need 2 Fe(OH)₃ molecules.
- O: There are 9 O atoms in 2 Fe(OH)₃ molecules (3 O atoms per molecule). On the right, there are 3 O atoms in Fe₂O₃ and 2 O atoms in 3 H₂O molecules. This is balanced.
- H: There are 6 H atoms in 2 Fe(OH)₃ molecules. On the right, there are 6 H atoms in 3 H₂O molecules. This is balanced.
Final balanced equation:
$$
2 \text{Fe(OH)}_3 \rightarrow \text{Fe}_2\text{O}_3 + 3 \text{H}_2\text{O}
$$
---
4) UO₃ + H₂ → UO₂ + H₂O
- Reactants: UO₃, H₂
- Products: UO₂, H₂O
#### Balancing:
- U: There is 1 U atom on both sides, so it is balanced.
- O: There are 3 O atoms in UO₃. On the right, there are 2 O atoms in UO₂ and 1 O atom in H₂O. To balance O, we need 2 H₂O molecules.
- H: There are 4 H atoms in 2 H₂ molecules. On the right, there are 4 H atoms in 2 H₂O molecules. This is balanced.
Final balanced equation:
$$
UO_3 + 2H_2 \rightarrow UO_2 + 2H_2O
$$
---
5) V₂O₅ + Al → Al₂O₃ + V
- Reactants: V₂O₅, Al
- Products: Al₂O₃, V
#### Balancing:
- V: There are 2 V atoms in V₂O₅, so we need 2 V atoms on the right.
- O: There are 5 O atoms in V₂O₅. On the right, there are 3 O atoms in Al₂O₃. To balance O, we need 5/3 Al₂O₃ molecules, but since we can't have fractional coefficients, multiply everything by 3.
- Al: There are 2 Al atoms in Al₂O₃, so we need 6 Al atoms on the left.
Final balanced equation:
$$
3V_2O_5 + 10Al \rightarrow 5Al_2O_3 + 6V
$$
---
6) PCl₅ + KNO₂ → NOCl + POCl₃ + KCl
- Reactants: PCl₅, KNO₂
- Products: NOCl, POCl₃, KCl
#### Balancing:
- P: There is 1 P atom in PCl₅, so we need 1 P atom in POCl₃.
- Cl: There are 5 Cl atoms in PCl₅. On the right, there is 1 Cl atom in NOCl, 3 Cl atoms in POCl₃, and 1 Cl atom in KCl. This is balanced.
- K: There is 1 K atom in KNO₂, so we need 1 K atom in KCl.
- N: There is 1 N atom in KNO₂, so we need 1 N atom in NOCl.
- O: There are 2 O atoms in KNO₂. On the right, there is 1 O atom in NOCl and 3 O atoms in POCl₃. This is balanced.
Final balanced equation:
$$
PCl_5 + KNO_2 \rightarrow NOCl + POCl_3 + KCl
$$
---
7) NaOH + Zn(NO₃)₂ → NaNO₃ + Zn(OH)₂
- Reactants: NaOH, Zn(NO₃)₂
- Products: NaNO₃, Zn(OH)₂
#### Balancing:
- Na: There is 1 Na atom in NaOH and 1 Na atom in NaNO₃, so we need 2 NaOH molecules.
- Zn: There is 1 Zn atom in Zn(NO₃)₂ and 1 Zn atom in Zn(OH)₂, so it is balanced.
- N: There are 2 N atoms in Zn(NO₃)₂ and 2 N atoms in 2 NaNO₃, so it is balanced.
- O: There are 6 O atoms in Zn(NO₃)₂ and 2 O atoms in 2 NaOH. On the right, there are 3 O atoms in 2 NaNO₃ and 2 O atoms in Zn(OH)₂. This is balanced.
- H: There are 2 H atoms in 2 NaOH and 2 H atoms in Zn(OH)₂. This is balanced.
Final balanced equation:
$$
2NaOH + Zn(NO_3)_2 \rightarrow 2NaNO_3 + Zn(OH)_2
$$
---
8) Bi(NO₃)₃ + H₂S → Bi₂S₃ + HNO₃
- Reactants: Bi(NO₃)₃, H₂S
- Products: Bi₂S₃, HNO₃
#### Balancing:
- Bi: There are 2 Bi atoms in Bi₂S₃, so we need 2 Bi(NO₃)₃ molecules.
- S: There are 3 S atoms in Bi₂S₃, so we need 3 H₂S molecules.
- N: There are 6 N atoms in 2 Bi(NO₃)₃, so we need 6 HNO₃ molecules.
- O: There are 18 O atoms in 2 Bi(NO₃)₃. On the right, there are 3 O atoms in Bi₂S₃ and 3 O atoms in each HNO₃ molecule. This is balanced.
- H: There are 6 H atoms in 3 H₂S molecules. On the right, there are 6 H atoms in 6 HNO₃ molecules. This is balanced.
Final balanced equation:
$$
2Bi(NO_3)_3 + 3H_2S \rightarrow Bi_2S_3 + 6HNO_3
$$
---
9) MgNH₄PO₄ → Mg₂P₂O₇ + NH₃ + H₂O
- Reactants: MgNH₄PO₄
- Products: Mg₂P₂O₇, NH₃, H₂O
#### Balancing:
- Mg: There are 2 Mg atoms in Mg₂P₂O₇, so we need 2 MgNH₄PO₄ molecules.
- P: There are 2 P atoms in Mg₂P₂O₇, so we need 2 MgNH₄PO₄ molecules.
- O: There are 7 O atoms in Mg₂P₂O₇. On the right, there are 4 O atoms in 2 H₂O molecules. This is balanced.
- N: There is 1 N atom in NH₃, so we need 2 NH₃ molecules.
- H: There are 8 H atoms in 2 MgNH₄PO₄ molecules. On the right, there are 6 H atoms in 2 NH₃ molecules and 4 H atoms in 2 H₂O molecules. This is balanced.
Final balanced equation:
$$
2MgNH_4PO_4 \rightarrow Mg_2P_2O_7 + 2NH_3 + 2H_2O
$$
---
10) Al + NH₄ClO₄ → Al₂O₃ + AlCl₃ + NO + H₂O
- Reactants: Al, NH₄ClO₄
- Products: Al₂O₃, AlCl₃, NO, H₂O
#### Balancing:
- Al: There are 2 Al atoms in Al₂O₃ and 3 Al atoms in AlCl₃, so we need 5 Al atoms on the left.
- Cl: There are 4 Cl atoms in NH₄ClO₄, so we need 4 AlCl₃ molecules.
- O: There are 4 O atoms in NH₄ClO₄. On the right, there are 3 O atoms in Al₂O₃, 1 O atom in NO, and 1 O atom in H₂O. To balance O, we need 6 H₂O molecules.
- N: There is 1 N atom in NH₄ClO₄, so we need 1 NO molecule.
- H: There are 4 H atoms in NH₄ClO₄. On the right, there are 2 H atoms in 2 H₂O molecules. To balance H, we need 2 NH₄ClO₄ molecules.
Final balanced equation:
$$
10Al + 6NH_4ClO_4 \rightarrow 2Al_2O_3 + 4AlCl_3 + 6NO + 12H_2O
$$
---
Final Answers:
1. $ 4 \text{Fe} + 3 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 $
2. $ P_4O_6 \rightarrow P_4 + 2P_2O_4 $
3. $ 2 \text{Fe(OH)}_3 \rightarrow \text{Fe}_2\text{O}_3 + 3 \text{H}_2\text{O} $
4. $ UO_3 + 2H_2 \rightarrow UO_2 + 2H_2O $
5. $ 3V_2O_5 + 10Al \rightarrow 5Al_2O_3 + 6V $
6. $ PCl_5 + KNO_2 \rightarrow NOCl + POCl_3 + KCl $
7. $ 2NaOH + Zn(NO_3)_2 \rightarrow 2NaNO_3 + Zn(OH)_2 $
8. $ 2Bi(NO_3)_3 + 3H_2S \rightarrow Bi_2S_3 + 6HNO_3 $
9. $ 2MgNH_4PO_4 \rightarrow Mg_2P_2O_7 + 2NH_3 + 2H_2O $
10. $ 10Al + 6NH_4ClO_4 \rightarrow 2Al_2O_3 + 4AlCl_3 + 6NO + 12H_2O $
$$
\boxed{
\begin{aligned}
&1. \ 4 \text{Fe} + 3 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 \\
&2. \ P_4O_6 \rightarrow P_4 + 2P_2O_4 \\
&3. \ 2 \text{Fe(OH)}_3 \rightarrow \text{Fe}_2\text{O}_3 + 3 \text{H}_2\text{O} \\
&4. \ UO_3 + 2H_2 \rightarrow UO_2 + 2H_2O \\
&5. \ 3V_2O_5 + 10Al \rightarrow 5Al_2O_3 + 6V \\
&6. \ PCl_5 + KNO_2 \rightarrow NOCl + POCl_3 + KCl \\
&7. \ 2NaOH + Zn(NO_3)_2 \rightarrow 2NaNO_3 + Zn(OH)_2 \\
&8. \ 2Bi(NO_3)_3 + 3H_2S \rightarrow Bi_2S_3 + 6HNO_3 \\
&9. \ 2MgNH_4PO_4 \rightarrow Mg_2P_2O_7 + 2NH_3 + 2H_2O \\
&10. \ 10Al + 6NH_4ClO_4 \rightarrow 2Al_2O_3 + 4AlCl_3 + 6NO + 12H_2O \\
\end{aligned}
}
$$
Parent Tip: Review the logic above to help your child master the concept of balancing equations worksheet chemistry.
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