Ammonia burns in oxygen as `2NH_(3)(g)+2.5O_(2)(g) to 2NO(g)+3H_(2)O(g)`
What volume of `O_(2)` at STP is required to produce 10 mol of products `(4NO+6H_(2)O)?`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction
The balanced chemical equation is:
\[ 2NH_3(g) + 2.5O_2(g) \rightarrow 2NO(g) + 3H_2O(g) \]

From this equation, we can see that:
- 2 moles of \( O_2 \) are required to produce 5 moles of products (2 moles of \( NO \) and 3 moles of \( H_2O \)).

### Step 2: Determine the Total Moles of Products
The problem states that we want to produce a total of 10 moles of products, which consists of:
- 4 moles of \( NO \)
- 6 moles of \( H_2O \)

The total moles of products is:
\[ 4 + 6 = 10 \text{ moles} \]

### Step 3: Calculate the Moles of \( O_2 \) Required
From the balanced equation, we know that:
- 5 moles of products require 2.5 moles of \( O_2 \).

To find out how many moles of \( O_2 \) are needed for 10 moles of products, we set up a proportion:
\[ \text{If } 5 \text{ moles of products require } 2.5 \text{ moles of } O_2, \text{ then } 10 \text{ moles of products require } x \text{ moles of } O_2. \]

Setting up the proportion:
\[ \frac{2.5 \text{ moles of } O_2}{5 \text{ moles of products}} = \frac{x \text{ moles of } O_2}{10 \text{ moles of products}} \]

Cross-multiplying gives:
\[ 2.5 \times 10 = 5 \times x \implies 25 = 5x \implies x = \frac{25}{5} = 5 \text{ moles of } O_2 \]

### Step 4: Convert Moles of \( O_2 \) to Volume at STP
At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, to find the volume of \( O_2 \):
\[ \text{Volume of } O_2 = 5 \text{ moles} \times 22.4 \text{ L/mole} = 112 \text{ liters} \]

### Final Answer
The volume of \( O_2 \) required to produce 10 moles of products is **112 liters**.

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