In the reaction, `4NH_(3)(g)+5O_(2)(g) rarr 4NO(g)+6H_(2)O(g)`, when 1 mole of ammonia and 1 mole of `O_(2)` are made to react to completion

Options

A. 1.0 mole of `H_(2)O` is produced
B. 1.0 mole of NO will be produced
C. all the oxygen will be consumed
D. all the ammonia will be consumed

Explanation

To solve the problem, we need to analyze the chemical reaction given and determine the limiting reagent when 1 mole of ammonia (NH₃) and 1 mole of oxygen (O₂) are reacted.

### Step 1: Write the balanced chemical equation
The balanced chemical equation is:
\[ 4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2O(g) \]

### Step 2: Determine the mole ratios from the balanced equation
From the balanced equation, we can see the following mole ratios:
- 4 moles of NH₃ react with 5 moles of O₂.
- This means that 1 mole of NH₃ requires \( \frac{5}{4} \) moles of O₂ to react completely.

### Step 3: Calculate the amount of O₂ required for 1 mole of NH₃
For 1 mole of NH₃:
\[ \text{O}_2 \text{ required} = \frac{5}{4} \text{ moles} = 1.25 \text{ moles} \]

### Step 4: Compare the available O₂ with the required O₂
We have only 1 mole of O₂ available, but we need 1.25 moles to completely react with 1 mole of NH₃. Therefore, O₂ is the limiting reagent.

### Step 5: Determine how much NH₃ will be consumed
Since O₂ is the limiting reagent, we need to calculate how much NH₃ will react with the available 1 mole of O₂.

Using the ratio from the balanced equation:
- 5 moles of O₂ produce 4 moles of NH₃.
- Therefore, 1 mole of O₂ will produce:
\[ \text{NH}_3 \text{ consumed} = \frac{4}{5} \text{ moles} = 0.8 \text{ moles} \]

### Step 6: Calculate the products formed
Using the limiting reagent (O₂), we can now calculate the amounts of NO and H₂O produced:
- From the balanced equation, 5 moles of O₂ produce 4 moles of NO.
- Therefore, 1 mole of O₂ will produce:
\[ \text{NO produced} = \frac{4}{5} \text{ moles} = 0.8 \text{ moles} \]

- From the balanced equation, 5 moles of O₂ produce 6 moles of H₂O.
- Therefore, 1 mole of O₂ will produce:
\[ \text{H}_2O \text{ produced} = \frac{6}{5} \text{ moles} = 1.2 \text{ moles} \]

### Summary of the results:
- Limiting reagent: O₂
- NH₃ consumed: 0.8 moles
- NO produced: 0.8 moles
- H₂O produced: 1.2 moles

### Final Conclusion:
In this reaction, 1 mole of NH₃ and 1 mole of O₂ react to completion, with O₂ being the limiting reagent. This results in the formation of 0.8 moles of NO and 1.2 moles of H₂O.

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