To balance the chemical equation for the combustion of ethane (C2H6), you need to ensure that the number of atoms for each element is the same on both sides of the equation. Here’s the step-by-step balancing process:
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Write the unbalanced equation: \[ C_2H_6 + O_2 \rightarrow CO_2 + H_2O \]
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Count the number of each type of atom on both sides:
- Reactants: 2 Carbon (C), 6 Hydrogen (H), and a variable amount of Oxygen (O)
- Products: 1 Carbon (C) in CO2, 2 Hydrogen (H) in H2O
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Start by balancing the carbon atoms: Since there are 2 carbon atoms in C2H6, we need 2 CO2: \[ C_2H_6 + O_2 \rightarrow 2CO_2 + H_2O \]
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Next, balance the hydrogen atoms: There are 6 hydrogen atoms in C2H6, so we need 3 H2O to have 6 hydrogen atoms: \[ C_2H_6 + O_2 \rightarrow 2CO_2 + 3H_2O \]
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Now count the oxygen atoms on the product side:
- From 2 CO2, there are 4 oxygen atoms.
- From 3 H2O, there are 3 oxygen atoms.
- Total = 4 + 3 = 7 oxygen atoms.
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On the reactant side, O2 has 2 oxygen atoms per molecule. To get 7 oxygen atoms, we need \( \frac{7}{2} \) O2, which is 3.5 O2. To avoid fractions, multiply the entire equation by 2: \[ 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O \]
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Now we count again:
- On the left: 4 Carbon (from 2 C2H6), 12 Hydrogen (from 2 C2H6), and 14 Oxygen (from 7 O2).
- On the right: 4 Carbon (from 4 CO2), 12 Hydrogen (from 6 H2O), and 14 Oxygen (4 CO2 contributes 8 + 6 from 3 H2O).
The balanced equation is: \[ \boxed{2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O} \]