Question
Balance the following chemical equations AICI3 + H2O = AL(OH)3 + HCI
Answers
GPT-4o mini
To balance the chemical equation \( \text{AlCl}_3 + \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + \text{HCl} \), follow these steps:
1. Write down the unbalanced equation:
\[
\text{AlCl}_3 + \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + \text{HCl}
\]
2. Count the number of each type of atom on both sides:
- **Reactants**:
- Al: 1 (from AlCl3)
- Cl: 3 (from AlCl3)
- H: 2 (from H2O)
- O: 1 (from H2O)
- **Products**:
- Al: 1 (from Al(OH)3)
- Cl: 1 (from HCl)
- H: 4 (3 from Al(OH)3 + 1 from HCl)
- O: 3 (from Al(OH)3)
3. Start balancing the equation:
- Balance the number of chlorine atoms by placing a coefficient of 3 in front of HCl:
\[
\text{AlCl}_3 + \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + 3 \text{HCl}
\]
4. Now recount and balance the hydrogen and oxygen:
- On the product side, we now have:
- H: 3 (from Al(OH)3) + 3 (from 3 HCl) = 6
- O: 3 (from Al(OH)3)
5. To balance the hydrogen, place a coefficient of 3 in front of water:
\[
\text{AlCl}_3 + 3 \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + 3 \text{HCl}
\]
6. Now recount the atoms:
- **Reactants**:
- Al: 1
- Cl: 3
- H: 6 (from 3 H2O)
- O: 3 (from 3 H2O)
- **Products**:
- Al: 1
- Cl: 3
- H: 6 (3 from Al(OH)3 and 3 from 3 HCl)
- O: 3 (from Al(OH)3)
Now everything is balanced. The final balanced equation is:
\[
\text{AlCl}_3 + 3 \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + 3 \text{HCl}
\]
1. Write down the unbalanced equation:
\[
\text{AlCl}_3 + \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + \text{HCl}
\]
2. Count the number of each type of atom on both sides:
- **Reactants**:
- Al: 1 (from AlCl3)
- Cl: 3 (from AlCl3)
- H: 2 (from H2O)
- O: 1 (from H2O)
- **Products**:
- Al: 1 (from Al(OH)3)
- Cl: 1 (from HCl)
- H: 4 (3 from Al(OH)3 + 1 from HCl)
- O: 3 (from Al(OH)3)
3. Start balancing the equation:
- Balance the number of chlorine atoms by placing a coefficient of 3 in front of HCl:
\[
\text{AlCl}_3 + \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + 3 \text{HCl}
\]
4. Now recount and balance the hydrogen and oxygen:
- On the product side, we now have:
- H: 3 (from Al(OH)3) + 3 (from 3 HCl) = 6
- O: 3 (from Al(OH)3)
5. To balance the hydrogen, place a coefficient of 3 in front of water:
\[
\text{AlCl}_3 + 3 \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + 3 \text{HCl}
\]
6. Now recount the atoms:
- **Reactants**:
- Al: 1
- Cl: 3
- H: 6 (from 3 H2O)
- O: 3 (from 3 H2O)
- **Products**:
- Al: 1
- Cl: 3
- H: 6 (3 from Al(OH)3 and 3 from 3 HCl)
- O: 3 (from Al(OH)3)
Now everything is balanced. The final balanced equation is:
\[
\text{AlCl}_3 + 3 \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + 3 \text{HCl}
\]
C3H8 + O2 = CO2 + H2O
GPT-4o mini
To balance the combustion reaction of propane (\( \text{C}_3\text{H}_8 \)), we start with the unbalanced equation:
\[
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
\]
### Steps to Balance the Equation:
1. **Count the Atoms**:
- **Reactants**:
- C: 3 (from \( \text{C}_3\text{H}_8 \))
- H: 8 (from \( \text{C}_3\text{H}_8 \))
- O: 2 (from \( \text{O}_2 \))
- **Products**:
- C: 1 (from \( \text{CO}_2 \))
- H: 2 (from \( \text{H}_2\text{O} \))
- O: 3 (2 from \( \text{CO}_2 \) and 1 from \( \text{H}_2\text{O} \))
2. **Balance Carbon Atoms**:
Place a coefficient of 3 in front of \( \text{CO}_2 \):
\[
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO}_2 + \text{H}_2\text{O}
\]
3. **New Count of Products**:
- **Products**:
- C: 3 (from 3 \( \text{CO}_2 \))
- H: 2 (from \( \text{H}_2\text{O} \))
- O: 7 (3 from 3 \( \text{CO}_2 \) and 1 from \( \text{H}_2\text{O} \))
4. **Balance Hydrogen Atoms**:
Place a coefficient of 4 in front of \( \text{H}_2\text{O} \):
\[
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O}
\]
5. **Recount the Atoms**:
- **Products**:
- C: 3
- H: 8 (4 from \( \text{H}_2\text{O} \))
- O: 10 (6 from 3 \( \text{CO}_2 \) and 4 from 4 \( \text{H}_2\text{O} \))
6. **Balance Oxygen Atoms**:
We now have 10 oxygen atoms in the products. Since \( O_2 \) has 2 oxygen atoms per molecule, we need 5 \( O_2 \) to provide 10 oxygen atoms:
\[
\text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O}
\]
### Final Balanced Equation:
\[
\text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O}
\]
This is the balanced equation for the combustion of propane.
\[
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
\]
### Steps to Balance the Equation:
1. **Count the Atoms**:
- **Reactants**:
- C: 3 (from \( \text{C}_3\text{H}_8 \))
- H: 8 (from \( \text{C}_3\text{H}_8 \))
- O: 2 (from \( \text{O}_2 \))
- **Products**:
- C: 1 (from \( \text{CO}_2 \))
- H: 2 (from \( \text{H}_2\text{O} \))
- O: 3 (2 from \( \text{CO}_2 \) and 1 from \( \text{H}_2\text{O} \))
2. **Balance Carbon Atoms**:
Place a coefficient of 3 in front of \( \text{CO}_2 \):
\[
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO}_2 + \text{H}_2\text{O}
\]
3. **New Count of Products**:
- **Products**:
- C: 3 (from 3 \( \text{CO}_2 \))
- H: 2 (from \( \text{H}_2\text{O} \))
- O: 7 (3 from 3 \( \text{CO}_2 \) and 1 from \( \text{H}_2\text{O} \))
4. **Balance Hydrogen Atoms**:
Place a coefficient of 4 in front of \( \text{H}_2\text{O} \):
\[
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O}
\]
5. **Recount the Atoms**:
- **Products**:
- C: 3
- H: 8 (4 from \( \text{H}_2\text{O} \))
- O: 10 (6 from 3 \( \text{CO}_2 \) and 4 from 4 \( \text{H}_2\text{O} \))
6. **Balance Oxygen Atoms**:
We now have 10 oxygen atoms in the products. Since \( O_2 \) has 2 oxygen atoms per molecule, we need 5 \( O_2 \) to provide 10 oxygen atoms:
\[
\text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O}
\]
### Final Balanced Equation:
\[
\text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O}
\]
This is the balanced equation for the combustion of propane.
AL + FeO = AI2O3 + Fe
GPT-4o mini
To balance the chemical equation \( \text{Al} + \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe} \), follow these steps:
1. **Write the unbalanced equation**:
\[
\text{Al} + \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}
\]
2. **Count the number of each type of atom on both sides**:
- **Reactants**:
- Al: 1
- Fe: 1
- O: 1
- **Products**:
- Al: 2 (from \( \text{Al}_2\text{O}_3 \))
- Fe: 1
- O: 3 (from \( \text{Al}_2\text{O}_3 \))
3. **Balance aluminum atoms**:
To balance the aluminum atoms, we need 2 Al in the reactants, so we place a coefficient of 2 in front of Al:
\[
2 \text{Al} + \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}
\]
4. **Recount the atoms**:
- **Reactants**:
- Al: 2
- Fe: 1
- O: 1
- **Products**:
- Al: 2
- Fe: 1
- O: 3
5. **Balance oxygen atoms**:
To balance the oxygen atoms, we know we need 3 oxygen atoms on the reactant side. Since each \( \text{FeO} \) provides 1 oxygen atom, we need 3 \( \text{FeO} \):
\[
2 \text{Al} + 3 \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}
\]
6. **Balance iron atoms**:
Now, with 3 \( \text{FeO} \), we have 3 iron atoms on the reactant side. Therefore, we should have a coefficient of 3 in front of \( \text{Fe} \) in the products:
\[
2 \text{Al} + 3 \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + 3 \text{Fe}
\]
7. **Final balanced equation**:
After ensuring all atoms are balanced, we get:
\[
2 \text{Al} + 3 \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + 3 \text{Fe}
\]
### Summary:
The balanced equation is:
\[
2 \text{Al} + 3 \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + 3 \text{Fe}
\]
1. **Write the unbalanced equation**:
\[
\text{Al} + \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}
\]
2. **Count the number of each type of atom on both sides**:
- **Reactants**:
- Al: 1
- Fe: 1
- O: 1
- **Products**:
- Al: 2 (from \( \text{Al}_2\text{O}_3 \))
- Fe: 1
- O: 3 (from \( \text{Al}_2\text{O}_3 \))
3. **Balance aluminum atoms**:
To balance the aluminum atoms, we need 2 Al in the reactants, so we place a coefficient of 2 in front of Al:
\[
2 \text{Al} + \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}
\]
4. **Recount the atoms**:
- **Reactants**:
- Al: 2
- Fe: 1
- O: 1
- **Products**:
- Al: 2
- Fe: 1
- O: 3
5. **Balance oxygen atoms**:
To balance the oxygen atoms, we know we need 3 oxygen atoms on the reactant side. Since each \( \text{FeO} \) provides 1 oxygen atom, we need 3 \( \text{FeO} \):
\[
2 \text{Al} + 3 \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}
\]
6. **Balance iron atoms**:
Now, with 3 \( \text{FeO} \), we have 3 iron atoms on the reactant side. Therefore, we should have a coefficient of 3 in front of \( \text{Fe} \) in the products:
\[
2 \text{Al} + 3 \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + 3 \text{Fe}
\]
7. **Final balanced equation**:
After ensuring all atoms are balanced, we get:
\[
2 \text{Al} + 3 \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + 3 \text{Fe}
\]
### Summary:
The balanced equation is:
\[
2 \text{Al} + 3 \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + 3 \text{Fe}
\]
S8 + O2 = SO3
GPT-4o mini
To balance the combustion reaction of sulfur (\( \text{S}_8 \)) with oxygen (\( \text{O}_2 \)) to form sulfur trioxide (\( \text{SO}_3 \)), follow these steps:
1. **Write the unbalanced equation**:
\[
\text{S}_8 + \text{O}_2 \rightarrow \text{SO}_3
\]
2. **Count the number of each type of atom on both sides**:
- **Reactants**:
- S: 8 (from \( \text{S}_8 \))
- O: 2 (in \( \text{O}_2 \))
- **Products**:
- S: 1 (from \( \text{SO}_3 \))
- O: 3 (from \( \text{SO}_3 \))
3. **Balance sulfur atoms**:
To balance the sulfur atoms, we need 8 sulfur atoms in the products. Therefore, place a coefficient of 8 in front of \( \text{SO}_3 \):
\[
\text{S}_8 + \text{O}_2 \rightarrow 8 \text{SO}_3
\]
4. **Recount the atoms**:
- **Products now**:
- S: 8 (from \( 8 \text{SO}_3 \))
- O: 24 (from \( 8 \text{SO}_3 \); since \( 8 \times 3 = 24 \))
5. **Balance oxygen atoms**:
We now have 24 oxygen atoms in the products and \( \text{O}_2 \) in the reactants, which means we need 12 \( \text{O}_2 \) molecules because each \( \text{O}_2 \) contains 2 oxygen atoms:
\[
\text{S}_8 + 12 \text{O}_2 \rightarrow 8 \text{SO}_3
\]
6. **Final balanced equation**:
After confirming all atoms balance, we have:
\[
\text{S}_8 + 12 \text{O}_2 \rightarrow 8 \text{SO}_3
\]
### Summary:
The balanced equation is:
\[
\text{S}_8 + 12 \text{O}_2 \rightarrow 8 \text{SO}_3
\]
1. **Write the unbalanced equation**:
\[
\text{S}_8 + \text{O}_2 \rightarrow \text{SO}_3
\]
2. **Count the number of each type of atom on both sides**:
- **Reactants**:
- S: 8 (from \( \text{S}_8 \))
- O: 2 (in \( \text{O}_2 \))
- **Products**:
- S: 1 (from \( \text{SO}_3 \))
- O: 3 (from \( \text{SO}_3 \))
3. **Balance sulfur atoms**:
To balance the sulfur atoms, we need 8 sulfur atoms in the products. Therefore, place a coefficient of 8 in front of \( \text{SO}_3 \):
\[
\text{S}_8 + \text{O}_2 \rightarrow 8 \text{SO}_3
\]
4. **Recount the atoms**:
- **Products now**:
- S: 8 (from \( 8 \text{SO}_3 \))
- O: 24 (from \( 8 \text{SO}_3 \); since \( 8 \times 3 = 24 \))
5. **Balance oxygen atoms**:
We now have 24 oxygen atoms in the products and \( \text{O}_2 \) in the reactants, which means we need 12 \( \text{O}_2 \) molecules because each \( \text{O}_2 \) contains 2 oxygen atoms:
\[
\text{S}_8 + 12 \text{O}_2 \rightarrow 8 \text{SO}_3
\]
6. **Final balanced equation**:
After confirming all atoms balance, we have:
\[
\text{S}_8 + 12 \text{O}_2 \rightarrow 8 \text{SO}_3
\]
### Summary:
The balanced equation is:
\[
\text{S}_8 + 12 \text{O}_2 \rightarrow 8 \text{SO}_3
\]
C7H6O2 + O2 = CO2 + H2O
GPT-4o mini
To balance the combustion reaction of benzoic acid (\( \text{C}_7\text{H}_6\text{O}_2 \)) with oxygen (\( \text{O}_2 \)), we start with the unbalanced equation:
\[
\text{C}_7\text{H}_6\text{O}_2 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
\]
### Steps to Balance the Equation:
1. **Count the Atoms in the Reactants**:
- Carbon (C): 7 (from \( \text{C}_7\text{H}_6\text{O}_2 \))
- Hydrogen (H): 6 (from \( \text{C}_7\text{H}_6\text{O}_2 \))
- Oxygen (O): 2 (from \( \text{C}_7\text{H}_6\text{O}_2 \)) + \( x \) (from \( \text{O}_2 \))
2. **Balance Carbon Atoms**:
Place a coefficient of 7 in front of \( \text{CO}_2 \):
\[
\text{C}_7\text{H}_6\text{O}_2 + \text{O}_2 \rightarrow 7 \text{CO}_2 + \text{H}_2\text{O}
\]
3. **Count the Atoms in the Products**:
- From \( 7 \text{CO}_2 \):
- Carbon (C): 7
- Oxygen (O): \( 7 \times 2 = 14 \) (from \( \text{CO}_2 \))
- Hydrogen (H): Needs to be balanced, so place a coefficient of 3 in front of \( \text{H}_2\text{O} \):
\[
\text{C}_7\text{H}_6\text{O}_2 + \text{O}_2 \rightarrow 7 \text{CO}_2 + 3 \text{H}_2\text{O}
\]
4. **Count the Atoms in the Products Again**:
- **Total in Products**:
- C: 7 (from \( 7 \text{CO}_2 \))
- H: 6 (from \( 3 \text{H}_2\text{O} \) since \( 3 \times 2 = 6 \))
- O: \( 14 + 3 = 17 \) (from \( 7 \text{CO}_2 \) and \( 3 \text{H}_2\text{O} \))
5. **Calculate the Needed Oxygen**:
The total number of oxygen atoms required on the reactant side is 17. The reactant \( \text{C}_7\text{H}_6\text{O}_2 \) already has 2 oxygen atoms, so we need \( 17 - 2 = 15 \) more oxygen atoms from \( \text{O}_2 \):
\[
15 \text{ O} / 2 = 7.5 \text{ O}_2
\]
6. **Modify for Whole Numbers**:
It's preferable to use whole numbers for coefficients. To eliminate the fraction, multiply the entire equation by 2:
\[
2 \text{C}_7\text{H}_6\text{O}_2 + 15 \text{O}_2 \rightarrow 14 \text{CO}_2 + 6 \text{H}_2\text{O}
\]
### Final Balanced Equation:
The final balanced equation for the combustion of benzoic acid is:
\[
2 \text{C}_7\text{H}_6\text{O}_2 + 15 \text{O}_2 \rightarrow 14 \text{CO}_2 + 6 \text{H}_2\text{O}
\]
\[
\text{C}_7\text{H}_6\text{O}_2 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
\]
### Steps to Balance the Equation:
1. **Count the Atoms in the Reactants**:
- Carbon (C): 7 (from \( \text{C}_7\text{H}_6\text{O}_2 \))
- Hydrogen (H): 6 (from \( \text{C}_7\text{H}_6\text{O}_2 \))
- Oxygen (O): 2 (from \( \text{C}_7\text{H}_6\text{O}_2 \)) + \( x \) (from \( \text{O}_2 \))
2. **Balance Carbon Atoms**:
Place a coefficient of 7 in front of \( \text{CO}_2 \):
\[
\text{C}_7\text{H}_6\text{O}_2 + \text{O}_2 \rightarrow 7 \text{CO}_2 + \text{H}_2\text{O}
\]
3. **Count the Atoms in the Products**:
- From \( 7 \text{CO}_2 \):
- Carbon (C): 7
- Oxygen (O): \( 7 \times 2 = 14 \) (from \( \text{CO}_2 \))
- Hydrogen (H): Needs to be balanced, so place a coefficient of 3 in front of \( \text{H}_2\text{O} \):
\[
\text{C}_7\text{H}_6\text{O}_2 + \text{O}_2 \rightarrow 7 \text{CO}_2 + 3 \text{H}_2\text{O}
\]
4. **Count the Atoms in the Products Again**:
- **Total in Products**:
- C: 7 (from \( 7 \text{CO}_2 \))
- H: 6 (from \( 3 \text{H}_2\text{O} \) since \( 3 \times 2 = 6 \))
- O: \( 14 + 3 = 17 \) (from \( 7 \text{CO}_2 \) and \( 3 \text{H}_2\text{O} \))
5. **Calculate the Needed Oxygen**:
The total number of oxygen atoms required on the reactant side is 17. The reactant \( \text{C}_7\text{H}_6\text{O}_2 \) already has 2 oxygen atoms, so we need \( 17 - 2 = 15 \) more oxygen atoms from \( \text{O}_2 \):
\[
15 \text{ O} / 2 = 7.5 \text{ O}_2
\]
6. **Modify for Whole Numbers**:
It's preferable to use whole numbers for coefficients. To eliminate the fraction, multiply the entire equation by 2:
\[
2 \text{C}_7\text{H}_6\text{O}_2 + 15 \text{O}_2 \rightarrow 14 \text{CO}_2 + 6 \text{H}_2\text{O}
\]
### Final Balanced Equation:
The final balanced equation for the combustion of benzoic acid is:
\[
2 \text{C}_7\text{H}_6\text{O}_2 + 15 \text{O}_2 \rightarrow 14 \text{CO}_2 + 6 \text{H}_2\text{O}
\]
BaF2 + K3PO4 = Ba3(PO4)2 +KF
GPT-4o mini
To balance the reaction between barium fluoride (\( \text{BaF}_2 \)) and potassium phosphate (\( \text{K}_3\text{PO}_4 \)) to form barium phosphate (\( \text{Ba}_3(\text{PO}_4)_2 \)) and potassium fluoride (\( \text{KF} \)), follow these steps:
1. **Write the unbalanced equation**:
\[
\text{BaF}_2 + \text{K}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + \text{KF}
\]
2. **Count the number of each type of atom on both sides**:
- **Reactants**:
- Ba: 1 (from \( \text{BaF}_2 \))
- F: 2 (from \( \text{BaF}_2 \))
- K: 3 (from \( \text{K}_3\text{PO}_4 \))
- P: 1 (from \( \text{K}_3\text{PO}_4 \))
- O: 4 (from \( \text{K}_3\text{PO}_4 \))
- **Products**:
- Ba: 3 (from \( \text{Ba}_3(\text{PO}_4)_2 \))
- F: 1 (from \( \text{KF} \))
- K: 1 (from \( \text{KF} \))
- P: 2 (from \( \text{Ba}_3(\text{PO}_4)_2 \))
- O: 8 (from \( \text{Ba}_3(\text{PO}_4)_2 \))
3. **Balance barium (Ba) first**:
To balance barium, put a coefficient of 3 in front of \( \text{BaF}_2 \):
\[
3\text{BaF}_2 + \text{K}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + \text{KF}
\]
4. **Recount the atoms**:
- **Reactants**:
- Ba: 3 (from \( 3\text{BaF}_2 \))
- F: 6 (from \( 3\text{BaF}_2 \))
- K: 3 (from \( \text{K}_3\text{PO}_4 \))
- P: 1 (from \( \text{K}_3\text{PO}_4 \))
- O: 4 (from \( \text{K}_3\text{PO}_4 \))
- **Products**:
- Ba: 3 (from \( \text{Ba}_3(\text{PO}_4)_2 \))
- F: 1 (from \( \text{KF} \))
- K: 1 (from \( \text{KF} \))
- P: 2 (from \( \text{Ba}_3(\text{PO}_4)_2 \))
- O: 8 (from \( \text{Ba}_3(\text{PO}_4)_2 \))
5. **Balance phosphate (P)**:
To account for the 2 phosphate groups in the product, we need to have 2 \( \text{K}_3\text{PO}_4 \):
\[
3\text{BaF}_2 + 2\text{K}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + \text{KF}
\]
6. **Balance water**:
As we balance potassium from \( \text{K}_3\text{PO}_4 \) (which contributes 6 K total), we need to balance that with \( \text{KF} \):
\[
3\text{BaF}_2 + 2\text{K}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + 6 \text{KF}
\]
### Final balanced equation:
After all adjustments, the fully balanced equation is:
\[
3\text{BaF}_2 + 2\text{K}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + 6\text{KF}
\]
### Summary:
The balanced reaction maintains the same number of atoms for each element on both sides of the equation.
1. **Write the unbalanced equation**:
\[
\text{BaF}_2 + \text{K}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + \text{KF}
\]
2. **Count the number of each type of atom on both sides**:
- **Reactants**:
- Ba: 1 (from \( \text{BaF}_2 \))
- F: 2 (from \( \text{BaF}_2 \))
- K: 3 (from \( \text{K}_3\text{PO}_4 \))
- P: 1 (from \( \text{K}_3\text{PO}_4 \))
- O: 4 (from \( \text{K}_3\text{PO}_4 \))
- **Products**:
- Ba: 3 (from \( \text{Ba}_3(\text{PO}_4)_2 \))
- F: 1 (from \( \text{KF} \))
- K: 1 (from \( \text{KF} \))
- P: 2 (from \( \text{Ba}_3(\text{PO}_4)_2 \))
- O: 8 (from \( \text{Ba}_3(\text{PO}_4)_2 \))
3. **Balance barium (Ba) first**:
To balance barium, put a coefficient of 3 in front of \( \text{BaF}_2 \):
\[
3\text{BaF}_2 + \text{K}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + \text{KF}
\]
4. **Recount the atoms**:
- **Reactants**:
- Ba: 3 (from \( 3\text{BaF}_2 \))
- F: 6 (from \( 3\text{BaF}_2 \))
- K: 3 (from \( \text{K}_3\text{PO}_4 \))
- P: 1 (from \( \text{K}_3\text{PO}_4 \))
- O: 4 (from \( \text{K}_3\text{PO}_4 \))
- **Products**:
- Ba: 3 (from \( \text{Ba}_3(\text{PO}_4)_2 \))
- F: 1 (from \( \text{KF} \))
- K: 1 (from \( \text{KF} \))
- P: 2 (from \( \text{Ba}_3(\text{PO}_4)_2 \))
- O: 8 (from \( \text{Ba}_3(\text{PO}_4)_2 \))
5. **Balance phosphate (P)**:
To account for the 2 phosphate groups in the product, we need to have 2 \( \text{K}_3\text{PO}_4 \):
\[
3\text{BaF}_2 + 2\text{K}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + \text{KF}
\]
6. **Balance water**:
As we balance potassium from \( \text{K}_3\text{PO}_4 \) (which contributes 6 K total), we need to balance that with \( \text{KF} \):
\[
3\text{BaF}_2 + 2\text{K}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + 6 \text{KF}
\]
### Final balanced equation:
After all adjustments, the fully balanced equation is:
\[
3\text{BaF}_2 + 2\text{K}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + 6\text{KF}
\]
### Summary:
The balanced reaction maintains the same number of atoms for each element on both sides of the equation.