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. (a)1.0 mole of H_(2)O is produced
B. (b)1.0 mole of NO will be produced
C. (c )All the oxygen will be consumed
D. (d)All the ammonia will be consumed

Explanation

To solve the problem, we need to analyze the given reaction and determine the limiting reagent when 1 mole of ammonia (NH₃) and 1 mole of oxygen (O₂) are reacted. Let's break it down step by step.

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction 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 ratio from the balanced equation
From the balanced equation, we can see that:
- 4 moles of NH₃ react with 5 moles of O₂.
- Therefore, the mole ratio of NH₃ to O₂ is 4:5.

### Step 3: Calculate the required amount of O₂ for 1 mole of NH₃
To find out how much O₂ is needed to react with 1 mole of NH₃, we can set up a proportion based on the mole ratio:
\[ \text{If } 4 \text{ moles of NH}_3 \text{ require } 5 \text{ moles of O}_2, \text{ then } 1 \text{ mole of NH}_3 \text{ requires } \frac{5}{4} \text{ moles of O}_2. \]
Calculating this gives:
\[ \frac{5}{4} = 1.25 \text{ moles of O}_2. \]

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

### Step 5: Calculate the amount of products formed
Now, we need to determine how much NO and H₂O will be produced from the 1 mole of O₂.

1. **For NO**:
According to the balanced equation, 5 moles of O₂ produce 4 moles of NO. Thus, 1 mole of O₂ will produce:
\[ \text{NO produced} = \frac{4}{5} \times 1 = 0.8 \text{ moles of NO}. \]

2. **For H₂O**:
According to the balanced equation, 5 moles of O₂ produce 6 moles of H₂O. Thus, 1 mole of O₂ will produce:
\[ \text{H}_2O \text{ produced} = \frac{6}{5} \times 1 = 1.2 \text{ moles of H}_2O. \]

### Step 6: Conclusion
- The limiting reagent is O₂.
- The amount of NO produced is 0.8 moles.
- The amount of H₂O produced is 1.2 moles.
- All of the O₂ will be consumed, but not all of the NH₃ will be consumed.