In the following reactions:
`4NH_(3)(g)+5O_(2)(g) to 4NO_(g)+6H_(2)O(l)`
When 1 mole ammonia and 1 mole of `O_(2)` are mixed, then the number of moles of NO formed will be:

Options

A. 0.8
B. 0.7
C. 0.6
D. 0.5

Explanation

To solve the problem, we need to analyze the given chemical reaction and determine how many moles of NO are produced when 1 mole of NH₃ and 1 mole of O₂ are mixed.

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

### Step 2: Identify the stoichiometric ratios.
From the balanced equation, we can see that:
- 4 moles of NH₃ react with 5 moles of O₂ to produce 4 moles of NO.

This gives us the following stoichiometric ratios:
- For NH₃ to NO: \( \frac{4 \text{ moles NO}}{4 \text{ moles NH}_3} = 1 \)
- For O₂ to NO: \( \frac{4 \text{ moles NO}}{5 \text{ moles O}_2} = 0.8 \)

### Step 3: Determine the limiting reactant.
We are mixing 1 mole of NH₃ and 1 mole of O₂. To find out which reactant is limiting, we can calculate how much NH₃ is required to react with 1 mole of O₂.

From the stoichiometric ratio, we know:
- 5 moles of O₂ require 4 moles of NH₃.
- Therefore, 1 mole of O₂ requires \( \frac{4}{5} \) moles of NH₃, which is 0.8 moles of NH₃.

Since we have 1 mole of NH₃ available, we can see that we have more than enough NH₃ to react with the available O₂. Thus, O₂ is the limiting reactant.

### Step 4: Calculate the amount of NO produced.
Now that we know O₂ is the limiting reactant, we can calculate how many moles of NO will be produced from the available O₂.

Using the stoichiometric ratio of O₂ to NO:
- 5 moles of O₂ produce 4 moles of NO.
- Therefore, 1 mole of O₂ will produce \( \frac{4}{5} \) moles of NO.

Calculating this gives:
\[ \text{Moles of NO} = 1 \text{ mole O}_2 \times \frac{4 \text{ moles NO}}{5 \text{ moles O}_2} = 0.8 \text{ moles NO} \]

### Final Answer:
The number of moles of NO formed when 1 mole of NH₃ and 1 mole of O₂ are mixed is **0.8 moles**.

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