Answer :
To balance the chemical equation for the reaction between [tex]\( FeCl_3(aq) \)[/tex] and [tex]\( Ba(OH)_2(aq) \)[/tex], we need to follow a systematic approach.
### Step-by-Step Solution:
1. Write down the reactants and products based on the reaction description:
Reactants:
[tex]\[ FeCl_3(aq) \][/tex]
[tex]\[ Ba(OH)_2(aq) \][/tex]
Products (based on double displacement reaction):
[tex]\[ Fe(OH)_3(s) \][/tex] (Solid product containing iron)
[tex]\[ BaCl_2(aq) \][/tex] (Aqueous product)
2. Write the unbalanced chemical equation, showing the double displacement:
[tex]\[ FeCl_3(aq) + Ba(OH)_2(aq) \rightarrow Fe(OH)_3(s) + BaCl_2(aq) \][/tex]
3. Identify the number of each type of atom on both sides of the equation:
- Left Side (Reactants):
- [tex]\( Fe: 1 \)[/tex]
- [tex]\( Cl: 3 \)[/tex]
- [tex]\( Ba: 1 \)[/tex]
- [tex]\( O: 2 \)[/tex]
- [tex]\( H: 2 \)[/tex]
- Right Side (Products):
- [tex]\( Fe: 1 \)[/tex]
- [tex]\( OH: 3 \)[/tex] (or [tex]\( O: 3 \)[/tex] and [tex]\( H: 3 \)[/tex])
- [tex]\( Ba: 1 \)[/tex]
- [tex]\( Cl: 2 \)[/tex]
4. Balance the chlorine atoms (Cl):
Since there are 3 chlorine atoms on the left side and only 2 on the right side, we need to balance the chlorine by increasing the number of [tex]\( BaCl_2 \)[/tex] molecules on the right side. To do this, multiply [tex]\( BaCl_2 \)[/tex] by 3:
[tex]\[ FeCl_3 + Ba(OH)_2 \rightarrow Fe(OH)_3 + 3BaCl_2 \][/tex]
5. Recalculate the number of atoms after increasing [tex]\( BaCl_2 \)[/tex] to 3:
- Left Side (Reactants):
- [tex]\( Fe: 1 \)[/tex]
- [tex]\( Cl: 3 \)[/tex]
- [tex]\( Ba: 1 \)[/tex]
- [tex]\( O: 2 \)[/tex]
- [tex]\( H: 2 \)[/tex]
- Right Side (Products):
- [tex]\( Fe: 1 \)[/tex]
- [tex]\( OH: 3 \)[/tex] (or [tex]\( O: 3 \)[/tex] and [tex]\( H: 3 \)[/tex]
- [tex]\( Ba: 3 \)[/tex]
- [tex]\( Cl: 6 \)[/tex]
6. Balance the barium atoms (Ba):
There are 3 barium atoms on the right side and only 1 on the left side. Balance the barium atoms by multiplying [tex]\( Ba(OH)_2 \)[/tex] by 3:
[tex]\[ FeCl_3 + 3Ba(OH)_2 \rightarrow Fe(OH)_3 + 3BaCl_2 \][/tex]
7. Recheck the number of atoms after making these adjustments:
- Left Side (Reactants):
- [tex]\( Fe: 1 \)[/tex]
- [tex]\( Cl: 3 \)[/tex]
- [tex]\( Ba: 3 \)[/tex]
- [tex]\( O: 6 \)[/tex]
- [tex]\( H: 6 \)[/tex]
- Right Side (Products):
- [tex]\( Fe: 1 \)[/tex]
- [tex]\( OH: 3 \)[/tex] (or [tex]\( O: 3 \)[/tex] and [tex]\( H: 3 \)[/tex]
- [tex]\( Ba: 3 \)[/tex]
- [tex]\( Cl: 6 \)[/tex]
Now, our equation is balanced in terms of all the atoms.
### Balanced Chemical Equation:
[tex]\[ FeCl_3(aq) + 3Ba(OH)_2(aq) \rightarrow Fe(OH)_3(s) + 3BaCl_2(aq) \][/tex]
This is the balanced chemical equation indicating that [tex]\(1\)[/tex] mole of iron(III) chloride reacts with [tex]\(3\)[/tex] moles of barium hydroxide to produce [tex]\(1\)[/tex] mole of iron(III) hydroxide and [tex]\(3\)[/tex] moles of barium chloride.
### Step-by-Step Solution:
1. Write down the reactants and products based on the reaction description:
Reactants:
[tex]\[ FeCl_3(aq) \][/tex]
[tex]\[ Ba(OH)_2(aq) \][/tex]
Products (based on double displacement reaction):
[tex]\[ Fe(OH)_3(s) \][/tex] (Solid product containing iron)
[tex]\[ BaCl_2(aq) \][/tex] (Aqueous product)
2. Write the unbalanced chemical equation, showing the double displacement:
[tex]\[ FeCl_3(aq) + Ba(OH)_2(aq) \rightarrow Fe(OH)_3(s) + BaCl_2(aq) \][/tex]
3. Identify the number of each type of atom on both sides of the equation:
- Left Side (Reactants):
- [tex]\( Fe: 1 \)[/tex]
- [tex]\( Cl: 3 \)[/tex]
- [tex]\( Ba: 1 \)[/tex]
- [tex]\( O: 2 \)[/tex]
- [tex]\( H: 2 \)[/tex]
- Right Side (Products):
- [tex]\( Fe: 1 \)[/tex]
- [tex]\( OH: 3 \)[/tex] (or [tex]\( O: 3 \)[/tex] and [tex]\( H: 3 \)[/tex])
- [tex]\( Ba: 1 \)[/tex]
- [tex]\( Cl: 2 \)[/tex]
4. Balance the chlorine atoms (Cl):
Since there are 3 chlorine atoms on the left side and only 2 on the right side, we need to balance the chlorine by increasing the number of [tex]\( BaCl_2 \)[/tex] molecules on the right side. To do this, multiply [tex]\( BaCl_2 \)[/tex] by 3:
[tex]\[ FeCl_3 + Ba(OH)_2 \rightarrow Fe(OH)_3 + 3BaCl_2 \][/tex]
5. Recalculate the number of atoms after increasing [tex]\( BaCl_2 \)[/tex] to 3:
- Left Side (Reactants):
- [tex]\( Fe: 1 \)[/tex]
- [tex]\( Cl: 3 \)[/tex]
- [tex]\( Ba: 1 \)[/tex]
- [tex]\( O: 2 \)[/tex]
- [tex]\( H: 2 \)[/tex]
- Right Side (Products):
- [tex]\( Fe: 1 \)[/tex]
- [tex]\( OH: 3 \)[/tex] (or [tex]\( O: 3 \)[/tex] and [tex]\( H: 3 \)[/tex]
- [tex]\( Ba: 3 \)[/tex]
- [tex]\( Cl: 6 \)[/tex]
6. Balance the barium atoms (Ba):
There are 3 barium atoms on the right side and only 1 on the left side. Balance the barium atoms by multiplying [tex]\( Ba(OH)_2 \)[/tex] by 3:
[tex]\[ FeCl_3 + 3Ba(OH)_2 \rightarrow Fe(OH)_3 + 3BaCl_2 \][/tex]
7. Recheck the number of atoms after making these adjustments:
- Left Side (Reactants):
- [tex]\( Fe: 1 \)[/tex]
- [tex]\( Cl: 3 \)[/tex]
- [tex]\( Ba: 3 \)[/tex]
- [tex]\( O: 6 \)[/tex]
- [tex]\( H: 6 \)[/tex]
- Right Side (Products):
- [tex]\( Fe: 1 \)[/tex]
- [tex]\( OH: 3 \)[/tex] (or [tex]\( O: 3 \)[/tex] and [tex]\( H: 3 \)[/tex]
- [tex]\( Ba: 3 \)[/tex]
- [tex]\( Cl: 6 \)[/tex]
Now, our equation is balanced in terms of all the atoms.
### Balanced Chemical Equation:
[tex]\[ FeCl_3(aq) + 3Ba(OH)_2(aq) \rightarrow Fe(OH)_3(s) + 3BaCl_2(aq) \][/tex]
This is the balanced chemical equation indicating that [tex]\(1\)[/tex] mole of iron(III) chloride reacts with [tex]\(3\)[/tex] moles of barium hydroxide to produce [tex]\(1\)[/tex] mole of iron(III) hydroxide and [tex]\(3\)[/tex] moles of barium chloride.
