When phosphours `(P_(4))` is heated in limited amount of `O_(2).P_(4)O_(6)` (tetraphosphorous hexaoxide) is obtained, and in excess of `O_(2)`, `P_(4) O_(10)` (tetraphosphours decaoxide) is obtained.
i. `P + 3 O_(2) rarr P_(4) O_(6)`, ii. `P_(4) + 5O_(2) rarr P_(4)O_(10)`
How many moles of `O_(2)` left unreacted initially in reaction (i) when 2g of `P_(4) reacts with 2g of `O_(2) ?

Options

A. 0.0145 mol
B. 0.072 mol
C. 0.029 mol
D. 0.0048 mol

Explanation

To solve the problem, we need to determine how many moles of \( O_2 \) are left unreacted after the reaction of \( P_4 \) with \( O_2 \) to form \( P_4O_6 \). Here are the steps to find the solution:

### Step 1: Calculate the moles of \( P_4 \)
The molar mass of phosphorus (\( P \)) is approximately 31 g/mol. Since \( P_4 \) consists of 4 phosphorus atoms, its molar mass is:
\[ \text{Molar mass of } P_4 = 4 \times 31 = 124 \, \text{g/mol} \]
Now, we can calculate the moles of \( P_4 \) using the formula:
\[ \text{Moles of } P_4 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{2 \, \text{g}}{124 \, \text{g/mol}} \approx 0.01613 \, \text{mol} \]

### Step 2: Calculate the moles of \( O_2 \)
The molar mass of oxygen (\( O_2 \)) is:
\[ \text{Molar mass of } O_2 = 2 \times 16 = 32 \, \text{g/mol} \]
Now, we can calculate the moles of \( O_2 \):
\[ \text{Moles of } O_2 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{2 \, \text{g}}{32 \, \text{g/mol}} = 0.0625 \, \text{mol} \]

### Step 3: Determine the stoichiometry of the reaction
From the reaction:
\[ P_4 + 3 O_2 \rightarrow P_4O_6 \]
We see that 1 mole of \( P_4 \) reacts with 3 moles of \( O_2 \). Therefore, the moles of \( O_2 \) required for the available moles of \( P_4 \) can be calculated as:
\[ \text{Moles of } O_2 \text{ required} = 3 \times \text{Moles of } P_4 = 3 \times 0.01613 \approx 0.04839 \, \text{mol} \]

### Step 4: Determine the limiting reagent
We have:
- Moles of \( P_4 \) = 0.01613 mol
- Moles of \( O_2 \) = 0.0625 mol
- Moles of \( O_2 \) required = 0.04839 mol

Since \( O_2 \) is in excess and \( P_4 \) is the limiting reagent, we will proceed to calculate the unreacted \( O_2 \).

### Step 5: Calculate the unreacted \( O_2 \)
To find the moles of \( O_2 \) left unreacted:
\[ \text{Moles of } O_2 \text{ left unreacted} = \text{Moles of } O_2 \text{ available} - \text{Moles of } O_2 \text{ reacted} \]
\[ = 0.0625 \, \text{mol} - 0.04839 \, \text{mol} \approx 0.01411 \, \text{mol} \]

### Final Answer
The number of moles of \( O_2 \) left unreacted is approximately \( 0.01411 \, \text{mol} \).