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)`
What mass of `P_(4) O_(6)` will be produced by the combustion of `2.0 g` of `P_(4)` with `2.0 g` of `O_(2)`.

Options

A. 0.0145 mol
B. 0.0072 mol
C. 0.029
D. 0.0048

Explanation

To solve the problem of how much \( P_4O_6 \) will be produced by the combustion of \( 2.0 \, g \) of \( P_4 \) with \( 2.0 \, g \) of \( O_2 \), we will follow these steps:

### Step 1: Determine the moles of \( P_4 \) and \( O_2 \)

**Molar Mass Calculation:**
- Molar mass of \( P_4 \):
\[ 4 \times 31 \, g/mol = 124 \, g/mol \]
- Molar mass of \( O_2 \):
\[ 2 \times 16 \, g/mol = 32 \, g/mol \]

**Moles Calculation:**
- Moles of \( P_4 \):
\[ \text{Moles of } P_4 = \frac{2.0 \, g}{124 \, g/mol} = 0.0161 \, mol \]
- Moles of \( O_2 \):
\[ \text{Moles of } O_2 = \frac{2.0 \, g}{32 \, g/mol} = 0.0625 \, mol \]

### Step 2: Identify the limiting reagent

From the reaction:
\[ P_4 + 3O_2 \rightarrow P_4O_6 \]
1 mole of \( P_4 \) reacts with 3 moles of \( O_2 \).

**Calculate required moles of \( O_2 \) for \( 0.0161 \, mol \) of \( P_4 \):**
\[ \text{Required } O_2 = 0.0161 \, mol \times 3 = 0.0483 \, mol \]

Since we have \( 0.0625 \, mol \) of \( O_2 \) available, \( O_2 \) is in excess, making \( P_4 \) the limiting reagent.

### Step 3: Calculate moles of \( P_4O_6 \) produced

From the stoichiometry of the reaction:
\[ 1 \, mol \, P_4 \rightarrow 1 \, mol \, P_4O_6 \]

Thus, \( 0.0161 \, mol \) of \( P_4 \) will produce:
\[ 0.0161 \, mol \, P_4O_6 \]

### Step 4: Calculate the mass of \( P_4O_6 \)

**Molar Mass of \( P_4O_6 \):**
\[ \text{Molar mass of } P_4O_6 = 4 \times 31 + 6 \times 16 = 124 + 96 = 220 \, g/mol \]

**Mass Calculation:**
\[ \text{Mass of } P_4O_6 = \text{Moles} \times \text{Molar Mass} = 0.0161 \, mol \times 220 \, g/mol = 3.542 \, g \]

### Final Answer
The mass of \( P_4O_6 \) produced is approximately **3.54 grams**.

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