The reaction you provided is a combustion reaction of butane (CH₃CH₂CH₂CH₃) with oxygen (O₂) producing carbon dioxide (CO₂) and water (H₂O). To balance this combustion reaction, we need to ensure that the number of each type of atom is the same on both sides of the equation.
The unbalanced equation is: \[ \text{C}4\text{H}{10} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]
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Count the number of atoms:
- Reactants: 4 carbon (C) atoms, 10 hydrogen (H) atoms
- Products:
- For every CO₂ produced, 1 C atom is used.
- For every H₂O produced, 2 H atoms are used.
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Set up the equation:
- Since there are 4 carbon atoms in butane, you need 4 CO₂: \[ \text{C}4\text{H}{10} + \text{O}_2 \rightarrow 4 \text{CO}_2 + \text{H}_2\text{O} \]
- The sum of hydrogen atoms in water requires 5 H₂O (since 10 H atoms in butane): \[ \text{C}4\text{H}{10} + \text{O}_2 \rightarrow 4 \text{CO}_2 + 5 \text{H}_2\text{O} \]
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Balance the oxygen atoms:
- On the right side, you have:
- 4 CO₂ contributes 4 × 2 = 8 O atoms.
- 5 H₂O contributes 5 × 1 = 5 O atoms.
- Total O atoms needed = 8 + 5 = 13 O atoms.
- Oxygen (O₂) has 2 O atoms per molecule; therefore, you need \(\frac{13}{2}\) O₂ molecules, or 6.5 O₂.
- On the right side, you have:
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Final balanced equation: To avoid fractional coefficients, multiply everything by 2: \[ 2 \text{C}4\text{H}{10} + 13 \text{O}_2 \rightarrow 8 \text{CO}_2 + 10 \text{H}_2\text{O} \]
This gives the balanced equation: \[ 2 \text{C}4\text{H}{10} + 13 \text{O}_2 \rightarrow 8 \text{CO}_2 + 10 \text{H}_2\text{O} \]
This equation shows the complete combustion of butane with oxygen.
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