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Certainly! To balance the chemical equation [tex]\(\text{CH}_3\text{CH}_3(g) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(g)\)[/tex], let's follow a systematic approach step by step.
1. Write down the unbalanced equation:
[tex]\[ \text{C}_2\text{H}_6(g) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(g) \][/tex]
2. Count the number of each type of atom on both sides:
- Reactants:
- Carbon (C): 2 (from [tex]\(\text{C}_2\text{H}_6\)[/tex])
- Hydrogen (H): 6 (from [tex]\(\text{C}_2\text{H}_6\)[/tex])
- Oxygen (O): 2 (from [tex]\(\text{O}_2\)[/tex])
- Products:
- Carbon (C): 1 (from [tex]\(\text{CO}_2\)[/tex])
- Hydrogen (H): 2 (from [tex]\(\text{H}_2\text{O}\)[/tex])
- Oxygen (O): 3 (1 from [tex]\(\text{CO}_2\)[/tex] and 1 from [tex]\(\text{H}_2\text{O}\)[/tex])
3. Balance the Carbon atoms first:
There are 2 Carbon atoms in the reactant [tex]\(\text{C}_2\text{H}_6\)[/tex] and 1 Carbon atom per [tex]\(\text{CO}_2\)[/tex] molecule on the product side. So, we need 2 [tex]\(\text{CO}_2\)[/tex] molecules on the product side.
[tex]\[ \text{C}_2\text{H}_6(g) + \text{O}_2(g) \rightarrow 2\text{CO}_2(g) + \text{H}_2\text{O}(g) \][/tex]
4. Balance the Hydrogen atoms next:
There are 6 Hydrogen atoms in the reactant [tex]\(\text{C}_2\text{H}_6\)[/tex] and 2 Hydrogen atoms per [tex]\(\text{H}_2\text{O}\)[/tex] molecule on the product side. So, we need 3 [tex]\(\text{H}_2\text{O}\)[/tex] molecules on the product side.
[tex]\[ \text{C}_2\text{H}_6(g) + \text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(g) \][/tex]
5. Balance the Oxygen atoms last:
- There are on the product side:
- 4 Oxygen atoms from 2 [tex]\(\text{CO}_2\)[/tex] molecules.
- 3 Oxygen atoms from 3 [tex]\(\text{H}_2\text{O}\)[/tex] molecules.
- Total: [tex]\(4 + 3 = 7\)[/tex] Oxygen atoms.
- On the reactant side, [tex]\(\text{O}_2\)[/tex] has 2 Oxygen atoms per molecule. To balance [tex]\(7\)[/tex] Oxygen atoms, we need [tex]\(7 / 2 = 3.5\)[/tex] [tex]\(\text{O}_2\)[/tex] molecules.
It's not ideal to have a fraction in our balanced equation. To get whole-number coefficients, we multiply all coefficients by 2:
[tex]\[ 2 \text{C}_2\text{H}_6(g) + 7 \text{O}_2(g) \rightarrow 4 \text{CO}_2(g) + 6 \text{H}_2\text{O}(g) \][/tex]
6. Verify the number of atoms:
- Reactants:
- Carbon (C): [tex]\(2 \times 2 = 4\)[/tex]
- Hydrogen (H): [tex]\(2 \times 6 = 12\)[/tex]
- Oxygen (O): [tex]\(7 \times 2 = 14\)[/tex]
- Products:
- Carbon (C): [tex]\(4 \times 1 = 4\)[/tex]
- Hydrogen (H): [tex]\(6 \times 2 = 12\)[/tex]
- Oxygen (O): [tex]\(4 \times 2 + 6 \times 1 = 8 + 6 = 14\)[/tex]
The equation is balanced.
So the balanced chemical equation is:
[tex]\[ 2 \text{C}_2\text{H}_6(g) + 7 \text{O}_2(g) \rightarrow 4 \text{CO}_2(g) + 6 \text{H}_2\text{O}(g) \][/tex]
1. Write down the unbalanced equation:
[tex]\[ \text{C}_2\text{H}_6(g) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(g) \][/tex]
2. Count the number of each type of atom on both sides:
- Reactants:
- Carbon (C): 2 (from [tex]\(\text{C}_2\text{H}_6\)[/tex])
- Hydrogen (H): 6 (from [tex]\(\text{C}_2\text{H}_6\)[/tex])
- Oxygen (O): 2 (from [tex]\(\text{O}_2\)[/tex])
- Products:
- Carbon (C): 1 (from [tex]\(\text{CO}_2\)[/tex])
- Hydrogen (H): 2 (from [tex]\(\text{H}_2\text{O}\)[/tex])
- Oxygen (O): 3 (1 from [tex]\(\text{CO}_2\)[/tex] and 1 from [tex]\(\text{H}_2\text{O}\)[/tex])
3. Balance the Carbon atoms first:
There are 2 Carbon atoms in the reactant [tex]\(\text{C}_2\text{H}_6\)[/tex] and 1 Carbon atom per [tex]\(\text{CO}_2\)[/tex] molecule on the product side. So, we need 2 [tex]\(\text{CO}_2\)[/tex] molecules on the product side.
[tex]\[ \text{C}_2\text{H}_6(g) + \text{O}_2(g) \rightarrow 2\text{CO}_2(g) + \text{H}_2\text{O}(g) \][/tex]
4. Balance the Hydrogen atoms next:
There are 6 Hydrogen atoms in the reactant [tex]\(\text{C}_2\text{H}_6\)[/tex] and 2 Hydrogen atoms per [tex]\(\text{H}_2\text{O}\)[/tex] molecule on the product side. So, we need 3 [tex]\(\text{H}_2\text{O}\)[/tex] molecules on the product side.
[tex]\[ \text{C}_2\text{H}_6(g) + \text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(g) \][/tex]
5. Balance the Oxygen atoms last:
- There are on the product side:
- 4 Oxygen atoms from 2 [tex]\(\text{CO}_2\)[/tex] molecules.
- 3 Oxygen atoms from 3 [tex]\(\text{H}_2\text{O}\)[/tex] molecules.
- Total: [tex]\(4 + 3 = 7\)[/tex] Oxygen atoms.
- On the reactant side, [tex]\(\text{O}_2\)[/tex] has 2 Oxygen atoms per molecule. To balance [tex]\(7\)[/tex] Oxygen atoms, we need [tex]\(7 / 2 = 3.5\)[/tex] [tex]\(\text{O}_2\)[/tex] molecules.
It's not ideal to have a fraction in our balanced equation. To get whole-number coefficients, we multiply all coefficients by 2:
[tex]\[ 2 \text{C}_2\text{H}_6(g) + 7 \text{O}_2(g) \rightarrow 4 \text{CO}_2(g) + 6 \text{H}_2\text{O}(g) \][/tex]
6. Verify the number of atoms:
- Reactants:
- Carbon (C): [tex]\(2 \times 2 = 4\)[/tex]
- Hydrogen (H): [tex]\(2 \times 6 = 12\)[/tex]
- Oxygen (O): [tex]\(7 \times 2 = 14\)[/tex]
- Products:
- Carbon (C): [tex]\(4 \times 1 = 4\)[/tex]
- Hydrogen (H): [tex]\(6 \times 2 = 12\)[/tex]
- Oxygen (O): [tex]\(4 \times 2 + 6 \times 1 = 8 + 6 = 14\)[/tex]
The equation is balanced.
So the balanced chemical equation is:
[tex]\[ 2 \text{C}_2\text{H}_6(g) + 7 \text{O}_2(g) \rightarrow 4 \text{CO}_2(g) + 6 \text{H}_2\text{O}(g) \][/tex]
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