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Sagot :
To balance the chemical equation:
[tex]\[ H_2CO_3 + Al(OH)_3 \rightarrow Al_2(CO_3)_3 + H_2O \][/tex]
we need to ensure that the number of atoms of each element on the reactants side is equal to the number of atoms on the products side.
Step-by-Step Solution:
1. Write down the initial number of atoms for each element on both sides of the equation:
Reactants:
- [tex]\( H_2CO_3 \)[/tex] : 2 Hydrogen (H), 1 Carbon (C), 3 Oxygen (O).
- [tex]\( Al(OH)_3 \)[/tex] : 1 Aluminum (Al), 3 Oxygen (O), 3 Hydrogen (H) per molecule.
Products:
- [tex]\( Al_2(CO_3)_3 \)[/tex]: 2 Aluminum (Al), 3 Carbon (C), 9 Oxygen (O).
- [tex]\( H_2O \)[/tex]: 2 Hydrogen (H), 1 Oxygen (O) per molecule.
2. Count the total atoms in the reactants and products:
Reactants:
- Hydrogen (H): [tex]\(2 + 3 \times 1 = 5\)[/tex]
- Carbon (C): [tex]\(1\)[/tex]
- Oxygen (O): [tex]\(3 + 3 \times 1 = 6\)[/tex]
- Aluminum (Al): [tex]\(1\)[/tex]
Products:
- Hydrogen (H): [tex]\(2 \times 1 = 2\)[/tex]
- Carbon (C): [tex]\(3\)[/tex]
- Oxygen (O): [tex]\(3 \times 3 + 1 = 10\)[/tex]
- Aluminum (Al): [tex]\(2\)[/tex]
We observe an imbalance in the atoms of hydrogen, carbon, oxygen, and aluminum.
3. Balance each element one by one:
- Balance Aluminum (Al): The products side has 2 Al atoms. Thus, we need 2 Al atoms on the reactants side.
[tex]\[ 2 \, Al(OH)_3 \rightarrow 2 \, Al + 6 \, OH \][/tex]
This means we have doubled the aluminum atoms.
- Balance Carbon (C): The products side has 3 C atoms. Thus, we need 3 H2CO3 molecules on the reactants side.
[tex]\[ 3 \, H2CO3 \rightarrow 3 \, C + 9 \, O + 6 \, H \][/tex]
- Recount atoms after adjustments:
- Reactants:
- Hydrogen (H): [tex]\(3 \times 2 + 2 \times 3 = 12\)[/tex]
- Carbon (C): [tex]\(3\)[/tex]
- Oxygen (O): [tex]\(3 \times 3 + 2 \times 3 = 15\)[/tex]
- Aluminum (Al): [tex]\(2\)[/tex]
- Products:
- Hydrogen (H): [tex]\(2 \times 3 = 6\)[/tex]
- Carbon (C): [tex]\(3\)[/tex]
- Oxygen (O): [tex]\(9 + 3 = 12\)[/tex]
- Aluminum (Al): [tex]\(2\)[/tex]
4. Balance Hydrogen and Oxygen in the products to reactants:
Since we balanced the Aluminum and Carbon correctly, we need to update the products side hydrogen and oxygen atom counts to match that of the reactants.
- Water (H2O): The number of hydrogen atoms on the reactants side is 12, which means we need 6 H2O molecules to balance it out:
[tex]\[ 3 H2O + 9 O (Balanced by oxygens from Al2(CO3)3 ) \][/tex]
5. The balanced chemical equation is:
[tex]\[ 3 \, H_2CO_3 + 2 \, Al(OH)_3 \rightarrow 1 \, Al_2(CO_3)_3 + 3 \, H_2O \][/tex]
To summarize, after balancing all atoms in the chemical equation, we obtain one mole consistent and balanced equation:
[tex]\[ 3 H_2CO_3 + 2 Al(OH)_3 \rightarrow Al_2(CO_3)_3 + 3 H_2O \][/tex]
[tex]\[ H_2CO_3 + Al(OH)_3 \rightarrow Al_2(CO_3)_3 + H_2O \][/tex]
we need to ensure that the number of atoms of each element on the reactants side is equal to the number of atoms on the products side.
Step-by-Step Solution:
1. Write down the initial number of atoms for each element on both sides of the equation:
Reactants:
- [tex]\( H_2CO_3 \)[/tex] : 2 Hydrogen (H), 1 Carbon (C), 3 Oxygen (O).
- [tex]\( Al(OH)_3 \)[/tex] : 1 Aluminum (Al), 3 Oxygen (O), 3 Hydrogen (H) per molecule.
Products:
- [tex]\( Al_2(CO_3)_3 \)[/tex]: 2 Aluminum (Al), 3 Carbon (C), 9 Oxygen (O).
- [tex]\( H_2O \)[/tex]: 2 Hydrogen (H), 1 Oxygen (O) per molecule.
2. Count the total atoms in the reactants and products:
Reactants:
- Hydrogen (H): [tex]\(2 + 3 \times 1 = 5\)[/tex]
- Carbon (C): [tex]\(1\)[/tex]
- Oxygen (O): [tex]\(3 + 3 \times 1 = 6\)[/tex]
- Aluminum (Al): [tex]\(1\)[/tex]
Products:
- Hydrogen (H): [tex]\(2 \times 1 = 2\)[/tex]
- Carbon (C): [tex]\(3\)[/tex]
- Oxygen (O): [tex]\(3 \times 3 + 1 = 10\)[/tex]
- Aluminum (Al): [tex]\(2\)[/tex]
We observe an imbalance in the atoms of hydrogen, carbon, oxygen, and aluminum.
3. Balance each element one by one:
- Balance Aluminum (Al): The products side has 2 Al atoms. Thus, we need 2 Al atoms on the reactants side.
[tex]\[ 2 \, Al(OH)_3 \rightarrow 2 \, Al + 6 \, OH \][/tex]
This means we have doubled the aluminum atoms.
- Balance Carbon (C): The products side has 3 C atoms. Thus, we need 3 H2CO3 molecules on the reactants side.
[tex]\[ 3 \, H2CO3 \rightarrow 3 \, C + 9 \, O + 6 \, H \][/tex]
- Recount atoms after adjustments:
- Reactants:
- Hydrogen (H): [tex]\(3 \times 2 + 2 \times 3 = 12\)[/tex]
- Carbon (C): [tex]\(3\)[/tex]
- Oxygen (O): [tex]\(3 \times 3 + 2 \times 3 = 15\)[/tex]
- Aluminum (Al): [tex]\(2\)[/tex]
- Products:
- Hydrogen (H): [tex]\(2 \times 3 = 6\)[/tex]
- Carbon (C): [tex]\(3\)[/tex]
- Oxygen (O): [tex]\(9 + 3 = 12\)[/tex]
- Aluminum (Al): [tex]\(2\)[/tex]
4. Balance Hydrogen and Oxygen in the products to reactants:
Since we balanced the Aluminum and Carbon correctly, we need to update the products side hydrogen and oxygen atom counts to match that of the reactants.
- Water (H2O): The number of hydrogen atoms on the reactants side is 12, which means we need 6 H2O molecules to balance it out:
[tex]\[ 3 H2O + 9 O (Balanced by oxygens from Al2(CO3)3 ) \][/tex]
5. The balanced chemical equation is:
[tex]\[ 3 \, H_2CO_3 + 2 \, Al(OH)_3 \rightarrow 1 \, Al_2(CO_3)_3 + 3 \, H_2O \][/tex]
To summarize, after balancing all atoms in the chemical equation, we obtain one mole consistent and balanced equation:
[tex]\[ 3 H_2CO_3 + 2 Al(OH)_3 \rightarrow Al_2(CO_3)_3 + 3 H_2O \][/tex]
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