- To balance the chemical equation \( \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \):
Let's start with the unbalanced equation:
\[ \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]
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There are 2 carbon (C) atoms in \( \text{C}_2\text{H}_6 \), so we need 2 \( \text{CO}_2 \):
\[ \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow 2 \text{CO}_2 + \text{H}_2\text{O} \]
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There are 6 hydrogen (H) atoms in \( \text{C}_2\text{H}_6 \), so we need 3 \( \text{H}_2\text{O} \):
\[ \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow 2 \text{CO}_2 + 3 \text{H}_2\text{O} \]
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Now we can count the oxygen (O) atoms on the right side. There are:
- From \( 2 \text{CO}_2 \): \( 2 \times 2 = 4 \) O atoms
- From \( 3 \text{H}_2\text{O} \): \( 3 \times 1 = 3 \) O atoms
Total = \( 4 + 3 = 7 \) O atoms.
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Since each \( \text{O}_2 \) molecule has 2 O atoms, we need \( \frac{7}{2} \) or 3.5 \( \text{O}_2 \):
\[ \text{C}_2\text{H}_6 + \frac{7}{2} \text{O}_2 \rightarrow 2 \text{CO}_2 + 3 \text{H}_2\text{O} \]
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To eliminate the fraction, we can multiply the entire equation by 2:
\[ 2 \text{C}_2\text{H}_6 + 7 \text{O}_2 \rightarrow 4 \text{CO}_2 + 6 \text{H}_2\text{O} \]
So the balanced equation is:
- Blank 1: 2
- Blank 2: 7
- Blank 3: 4
- Blank 4: 6
- For the statement comparing synthesis and decomposition reactions, the correct option is:
b Synthesis and decomposition reactions are opposites of each other. Synthesis reactions are in the form A + B → AB, while decomposition reactions are in the form AB → A + B.