What weights of `P^(4)O_(6)` and `P_(4)O_(10)` will be produced by the combusion of 31g of `P_(4)` in 32g of oxygen leaving no `P_(4)` and `O_(2)` ?

Options

A. 2.75g, 219.5g
B. 27.5g, 35.5g
C. 55g, 71g
D. 17.5g, 190.5g

Answer

To solve the problem of determining the weights of \( P_4O_6 \) and \( P_4O_{10} \) produced by the combustion of 31 g of \( P_4 \) in 32 g of oxygen, we will follow these steps:

### Step 1: Write the balanced chemical equations for the combustion of \( P_4 \)
The combustion of phosphorus can produce two different oxides: \( P_4O_6 \) and \( P_4O_{10} \). The balanced equations are:

1. For \( P_4O_6 \):
\[ P_4 + 3O_2 \rightarrow 2P_4O_6 \]

2. For \( P_4O_{10} \):
\[ P_4 + 5O_2 \rightarrow P_4O_{10} \]

### Step 2: Calculate the moles of \( P_4 \) and \( O_2 \)
- Molar mass of \( P_4 \) = \( 4 \times 31 = 124 \, g/mol \)
- Molar mass of \( O_2 \) = \( 32 \, g/mol \)

**Moles of \( P_4 \)**:
\[ \text{Moles of } P_4 = \frac{31 \, g}{124 \, g/mol} = 0.25 \, mol \]

**Moles of \( O_2 \)**:
\[ \text{Moles of } O_2 = \frac{32 \, g}{32 \, g/mol} = 1 \, mol \]

### Step 3: Determine the limiting reagent
From the balanced equations:
- For \( P_4O_6 \): 1 mole of \( P_4 \) reacts with 3 moles of \( O_2 \).
- For \( P_4O_{10} \): 1 mole of \( P_4 \) reacts with 5 moles of \( O_2 \).

**Using \( P_4O_6 \)**:
- Required \( O_2 \) for \( 0.25 \, mol \) of \( P_4 \):
\[ 0.25 \, mol \times 3 = 0.75 \, mol \, O_2 \]

**Using \( P_4O_{10} \)**:
- Required \( O_2 \) for \( 0.25 \, mol \) of \( P_4 \):
\[ 0.25 \, mol \times 5 = 1.25 \, mol \, O_2 \]

Since we only have 1 mole of \( O_2 \), \( P_4O_6 \) will be produced as it requires less oxygen.

### Step 4: Calculate the amount of \( P_4O_6 \) produced
From the balanced equation for \( P_4O_6 \):
- 1 mole of \( P_4 \) produces 2 moles of \( P_4O_6 \).
- Therefore, \( 0.25 \, mol \) of \( P_4 \) will produce:
\[ 0.25 \, mol \times 2 = 0.5 \, mol \, P_4O_6 \]

**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 of \( P_4O_6 \)**:
\[ \text{Mass of } P_4O_6 = 0.5 \, mol \times 220 \, g/mol = 110 \, g \]

### Step 5: Calculate the amount of \( P_4O_{10} \) produced
Since all \( O_2 \) is consumed in the formation of \( P_4O_6 \), no \( P_4O_{10} \) will be produced.

### Final Answer
- Weight of \( P_4O_6 \): 110 g
- Weight of \( P_4O_{10} \): 0 g