To write and balance the chemical equation for the reaction of ammonia gas (NH₃) with oxygen gas (O₂) to produce nitrogen monoxide (NO) and water (H₂O), we can follow these steps:
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Write the unbalanced equation: \[ \text{NH}_3 (g) + \text{O}_2 (g) \rightarrow \text{NO} (g) + \text{H}_2\text{O} (g) \]
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Count the number of each type of atom on both sides:
- Reactants:
- N: 1 (from NH₃)
- H: 3 (from NH₃)
- O: 2 (from O₂)
- Products:
- N: 1 (from NO)
- H: 2 (from H₂O)
- O: 3 (1 from NO and 2 from H₂O)
- Reactants:
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Balance the equation:
- Start by balancing nitrogen atoms. There's 1 nitrogen atom on each side.
- Next, balance the hydrogen atoms. We have 3 H on the left, so we can place a coefficient of 3 in front of H₂O: \[ \text{NH}_3 (g) + \text{O}_2 (g) \rightarrow \text{NO} (g) + 3 \text{H}_2\text{O} (g) \]
Now recount the number of atoms:
- Reactants:
- N: 1
- H: 3
- O: 2
- Products:
- N: 1
- H: 6 (from 3 H₂O)
- O: 7 (1 from NO and 6 from 3 H₂O)
To balance the hydrogen atoms, place a coefficient of 3 in front of NH₃: \[ 3 \text{NH}_3 (g) + \text{O}_2 (g) \rightarrow 3 \text{NO} (g) + 3 \text{H}_2\text{O} (g) \]
Recount:
- Reactants:
- N: 3
- H: 9 (from 3 NH₃)
- O: 2 (from O₂)
- Products:
- N: 3 (from 3 NO)
- H: 6 (from 3 H₂O)
- O: 6 (from 3 H₂O) + 3 (from 3 NO) = 9
Now add. There are 3 O₂ + 6 O from H₂O.
This requires a coefficient of 5. Rewrite: \[ 4 \text{NH}_3 (g) + 5 \text{O}_2 (g) \rightarrow 4 \text{NO} (g) + 6 \text{H}_2\text{O} (g) \]
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Final balanced equation: \[ 4 \text{NH}_3 (g) + 5 \text{O}_2 (g) \rightarrow 4 \text{NO} (g) + 6 \text{H}_2\text{O} (g) \]
Now the equation is balanced, with equal numbers of each type of atom on both sides.