To balance the chemical equation \( \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \), we need to make sure that the number of atoms of each element is the same on both sides of the equation.
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Count the atoms of each element in the reactants:
- In \( \text{C}_2\text{H}_6 \): 2 carbon (C) atoms and 6 hydrogen (H) atoms.
- In \( \text{O}_2 \): 2 oxygen (O) atoms.
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Count the atoms of each element in the products:
- In \( \text{CO}_2 \): 1 carbon (C) atom and 2 oxygen (O) atoms.
- In \( \text{H}_2\text{O} \): 2 hydrogen (H) atoms and 1 oxygen (O) atom.
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Start balancing:
- Since there are 2 carbon atoms in the reactants (\( \text{C}_2\text{H}_6 \)), we need 2 \( \text{CO}_2 \) in the products. So we put a coefficient of 2 in front of \( \text{CO}_2 \): \[ \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow 2 \text{CO}_2 + \text{H}_2\text{O} \]
- Now we have 2 carbon atoms on both sides.
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Next, balance the hydrogen atoms:
- There are 6 hydrogen atoms in \( \text{C}_2\text{H}_6 \), so we need 3 \( \text{H}_2\text{O} \) in the products to get 6 hydrogen atoms: \[ \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow 2 \text{CO}_2 + 3 \text{H}_2\text{O} \]
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Count the oxygen atoms in the products:
- Now we have:
- From 2 \( \text{CO}_2 \): \( 2 \times 2 = 4 \) oxygen atoms.
- From 3 \( \text{H}_2\text{O} \): \( 3 \times 1 = 3 \) oxygen atoms.
- Total = \( 4 + 3 = 7 \) oxygen atoms in products.
- Now we have:
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Balance the oxygen atoms:
- Since \( \text{O}_2 \) gives us 2 oxygen atoms per molecule, we need \( \frac{7}{2} = 3.5 \) \( \text{O}_2 \) molecules.
- To avoid a fraction, multiply the entire equation by 2: \[ 2 \text{C}_2\text{H}_6 + 7 \text{O}_2 \rightarrow 4 \text{CO}_2 + 6 \text{H}_2\text{O} \]
Now, we can fill in the blanks:
Blank 1: 2
Blank 2: 7
Blank 3: 4
Blank 4: 6
So the balanced equation is: \[ 2 \text{C}_2\text{H}_6 + 7 \text{O}_2 \rightarrow 4 \text{CO}_2 + 6 \text{H}_2\text{O} \]
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