To balance the chemical equation \( C_2H_6 + O_2 \rightarrow CO_2 + H_2O \), we will follow these steps:
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Count the number of atoms of each element on both sides of the equation:
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On the left side:
- Carbon (C): 2 (from C2H6)
- Hydrogen (H): 6 (from C2H6)
- Oxygen (O): 2 (from O2)
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On the right side:
- Carbon (C): 1 (from CO2)
- Hydrogen (H): 2 (from H2O)
- Oxygen (O): 3 (1 from CO2 and 1 from H2O)
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Start balancing with carbon (C):
- There are 2 carbons on the left, so we need 2 CO2 on the right. \[ C_2H_6 + O_2 \rightarrow 2CO_2 + H_2O \]
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Now the equation looks like this:
- On the right side now we have:
- C: 2
- H: 6 (still needs balancing)
- O: 4 (from 2 CO2) + 1 (from H2O) = 5
- On the right side now we have:
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Next balance the hydrogen (H):
- There are 6 hydrogens in C2H6, so we need 3 H2O on the right. \[ C_2H_6 + O_2 \rightarrow 2CO_2 + 3H_2O \]
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Now count the atoms again:
- Right side:
- C: 2 (from 2CO2)
- H: 6 (from 3H2O)
- O: 4 (from 2CO2) + 3 (from 3H2O) = 7
- Right side:
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Now balance the oxygen (O):
- We have 7 oxygen atoms on the right side. In O2, each molecule contains 2 oxygen atoms, so we need: \[ \frac{7}{2} \text{ O}_2 = 3.5 \text{ O}_2 \] But it’s not preferred to have a fraction in a balanced equation, so we multiply everything by 2 to eliminate the fraction: \[ 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O \]
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Final balanced equation: \[ \boxed{2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O} \]
This balanced equation has equal numbers of each type of atom on both sides.