The half-life for the reaction `N_(2)O_(5) to 2 NO_(2) + (1)/(2) O_(2)` is 2.4 h at STP .
Starting with 10.8 g of `N_(2)O_(5)` how much oxygen will be obtained after a period of 9.6 h

Options

A. 1.5 L
B. 3.36 L
C. 1.05 L
D. 0.07 L

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the moles of N₂O₅
The molar mass of N₂O₅ (Nitrogen pentoxide) is calculated as follows:
- Nitrogen (N) = 14 g/mol, and there are 2 nitrogen atoms: 2 × 14 = 28 g/mol
- Oxygen (O) = 16 g/mol, and there are 5 oxygen atoms: 5 × 16 = 80 g/mol
- Therefore, the molar mass of N₂O₅ = 28 + 80 = 108 g/mol

Now, we can calculate the number of moles of N₂O₅ in 10.8 g:
\[ \text{Moles of } N₂O₅ = \frac{\text{mass}}{\text{molar mass}} = \frac{10.8 \text{ g}}{108 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 2: Determine the number of half-lives in 9.6 hours
The half-life (T₁/₂) of the reaction is given as 2.4 hours. To find the number of half-lives (N) that fit into 9.6 hours:
\[ N = \frac{\text{Total time}}{T_{1/2}} = \frac{9.6 \text{ hours}}{2.4 \text{ hours}} = 4 \]

### Step 3: Calculate the remaining moles of N₂O₅ after 4 half-lives
After each half-life, the amount of N₂O₅ remaining can be calculated as follows:
- After 1st half-life: \( \frac{0.1}{2} = 0.05 \) moles
- After 2nd half-life: \( \frac{0.05}{2} = 0.025 \) moles
- After 3rd half-life: \( \frac{0.025}{2} = 0.0125 \) moles
- After 4th half-life: \( \frac{0.0125}{2} = 0.00625 \) moles

Thus, the remaining moles of N₂O₅ after 9.6 hours is 0.00625 moles.

### Step 4: Calculate the moles of N₂O₅ that have reacted
The initial moles of N₂O₅ were 0.1 moles. The moles that have reacted can be calculated as:
\[ \text{Moles reacted} = \text{Initial moles} - \text{Remaining moles} = 0.1 - 0.00625 = 0.09375 \text{ moles} \]

### Step 5: Calculate the moles of O₂ produced
From the balanced reaction, 1 mole of N₂O₅ produces 0.5 moles of O₂. Therefore, the moles of O₂ produced from the reacted N₂O₅ can be calculated as:
\[ \text{Moles of O₂} = 0.5 \times \text{Moles reacted} = 0.5 \times 0.09375 = 0.046875 \text{ moles} \]

### Step 6: Calculate the volume of O₂ at STP
At Standard Temperature and Pressure (STP), 1 mole of gas occupies 22.4 liters. Therefore, the volume of O₂ produced is:
\[ \text{Volume of O₂} = \text{Moles of O₂} \times 22.4 \text{ L/mol} = 0.046875 \times 22.4 \approx 1.05 \text{ L} \]

### Final Answer
The volume of oxygen obtained after 9.6 hours is approximately **1.05 liters**.

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