To balance the skeleton equation \( \text{N}_2 + \text{H}_2\text{O} \rightarrow \text{NH}_3 + \text{NO} \), let's determine the coefficients needed for each compound:
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Write the unbalanced reaction: \[ \text{N}_2 + \text{H}_2\text{O} \rightarrow \text{NH}_3 + \text{NO} \]
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Count the number of each type of atom on both sides of the equation:
- Reactants:
- N: 2 (from \(\text{N}_2\))
- H: 2 (from \(\text{H}_2\text{O}\))
- O: 1 (from \(\text{H}_2\text{O}\))
- Products:
- N: 1 (from \(\text{NH}_3\)) + 1 (from \(\text{NO}\)) = 2
- H: 3 (from \(\text{NH}_3\))
- O: 1 (from \(\text{NO}\))
- Reactants:
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Balance the atoms:
- Nitrogen: Already balanced with 2 on each side.
- Hydrogen: Need to balance H by adjusting the coefficient in front of \(\text{H}_2\text{O}\):
- Since \(\text{NH}_3\) has 3 H atoms, we can adjust coefficients as follows: \[ \text{N}_2 + 3 \text{H}_2\text{O} \rightarrow 2 \text{NH}_3 + \text{NO} \]
- Now check again:
- Reactants:
- Nitrogen: 2
- Hydrogen: \(3 \times 2 = 6\)
- Oxygen: \(3 \times 1 = 3\)
- Products:
- Nitrogen: \(2 \text{NH}_3\) contributes 2 N and 1 N from \(\text{NO}\) = 3 N
- Hydrogen: \(2 \text{NH}_3\) contributes \(2 \times 3 = 6\)
- Oxygen: 1 from \(\text{NO}\)
- Reactants:
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Final balancing:
- The correct coefficients are now 1 for N2, 3 for H2O, 2 for NH3, and 1 for NO, hence the balanced equation is: \[ 1 \text{N}_2 + 3 \text{H}_2\text{O} \rightarrow 2 \text{NH}_3 + 1 \text{NO} \]
Therefore, the coefficients are 1, 3, 2, 1.
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