Two substances `P_(4)` &`O_(2)` are allowed to react completely to form mixture of `P_(4)O_(6)` &`P_(4)O_(10)` leaving none of the reactants .Using this information calculate the moles of `P_(4)O_(6)` and `P_(4)O_(10)` in the final mixture when the following amounts of `P_(4)` & `O_(2)` are taken.
`P_(4) + 3O_(2) to P_(4)O_(6)`
`P_(4) + 5O_(2) to P_(4)O_(10)`
(i) If 1 mole `P_(4)` & 4 mole of `O_(2)`
(ii) If 3 mole `P_(4)` & 11 mole of `O_(2)`
(iii) If 3 mole `P_(4)` & 13 mole of `O_(2)`

Answer

To solve the problem, we need to analyze the reactions of \( P_4 \) and \( O_2 \) to form \( P_4O_6 \) and \( P_4O_{10} \). We will use stoichiometry to determine the moles of products formed under different conditions.

### Given Reactions:
1. \( P_4 + 3O_2 \rightarrow P_4O_6 \)
2. \( P_4 + 5O_2 \rightarrow P_4O_{10} \)

### Conditions:
1. **(i)** If 1 mole \( P_4 \) & 4 moles \( O_2 \)
2. **(ii)** If 3 moles \( P_4 \) & 11 moles \( O_2 \)
3. **(iii)** If 3 moles \( P_4 \) & 13 moles \( O_2 \)

### Step-by-Step Solution:

#### (i) 1 mole \( P_4 \) & 4 moles \( O_2 \)

1. **Determine the limiting reactant:**
- For \( P_4O_6 \): 1 mole of \( P_4 \) requires 3 moles of \( O_2 \).
- For \( P_4O_{10} \): 1 mole of \( P_4 \) requires 5 moles of \( O_2 \).

Since we have 4 moles of \( O_2 \), we can use it for both reactions.

2. **Calculate moles of \( P_4O_6 \) produced:**
- Using 1 mole of \( P_4 \) and 3 moles of \( O_2 \) for \( P_4O_6 \):
\[ \text{Moles of } P_4O_6 = 1 \text{ mole } P_4 \text{ (used)} \rightarrow 1 \text{ mole } P_4O_6 \]

3. **Calculate remaining \( O_2 \):**
- After forming \( P_4O_6 \), we have:
\[ 4 \text{ moles } O_2 - 3 \text{ moles } O_2 = 1 \text{ mole } O_2 \]

4. **Use remaining \( O_2 \) for \( P_4O_{10} \):**
- 1 mole of \( O_2 \) is not enough to react with \( P_4 \) for \( P_4O_{10} \) (requires 5 moles).

5. **Final result:**
- Moles of \( P_4O_6 = 1 \)
- Moles of \( P_4O_{10} = 0 \)

#### (ii) 3 moles \( P_4 \) & 11 moles \( O_2 \)

1. **Determine the limiting reactant:**
- For \( P_4O_6 \): 3 moles of \( P_4 \) require \( 3 \times 3 = 9 \) moles of \( O_2 \).
- For \( P_4O_{10} \): 1 mole of \( P_4 \) requires 5 moles of \( O_2 \).

2. **Use \( P_4O_6 \) first:**
- Use 3 moles \( P_4 \) and 9 moles \( O_2 \):
\[ \text{Moles of } P_4O_6 = 3 \text{ moles } P_4 \rightarrow 3 \text{ moles } P_4O_6 \]

3. **Remaining \( O_2 \):**
\[ 11 \text{ moles } O_2 - 9 \text{ moles } O_2 = 2 \text{ moles } O_2 \]

4. **Use remaining \( O_2 \) for \( P_4O_{10} \):**
- 2 moles of \( O_2 \) are not enough to react with \( P_4 \) for \( P_4O_{10} \) (requires 5 moles).

5. **Final result:**
- Moles of \( P_4O_6 = 3 \)
- Moles of \( P_4O_{10} = 0 \)

#### (iii) 3 moles \( P_4 \) & 13 moles \( O_2 \)

1. **Determine the limiting reactant:**
- For \( P_4O_6 \): 3 moles of \( P_4 \) require 9 moles of \( O_2 \).
- For \( P_4O_{10} \): 1 mole of \( P_4 \) requires 5 moles of \( O_2 \).

2. **Use \( P_4O_6 \) first:**
- Use 3 moles \( P_4 \) and 9 moles \( O_2 \):
\[ \text{Moles of } P_4O_6 = 3 \text{ moles } P_4 \rightarrow 3 \text{ moles } P_4O_6 \]

3. **Remaining \( O_2 \):**
\[ 13 \text{ moles } O_2 - 9 \text{ moles } O_2 = 4 \text{ moles } O_2 \]

4. **Use remaining \( O_2 \) for \( P_4O_{10} \):**
- 4 moles of \( O_2 \) can react with \( P_4 \):
\[ 4 \text{ moles } O_2 \rightarrow \text{can react with } 1 \text{ mole } P_4 \]
- Therefore, we can form:
\[ \text{Moles of } P_4O_{10} = 1 \text{ mole } P_4 \rightarrow 1 \text{ mole } P_4O_{10} \]

5. **Final result:**
- Moles of \( P_4O_6 = 3 \)
- Moles of \( P_4O_{10} = 1 \)

### Summary of Results:
- (i) \( P_4O_6 = 1 \), \( P_4O_{10} = 0 \)
- (ii) \( P_4O_6 = 3 \), \( P_4O_{10} = 0 \)
- (iii) \( P_4O_6 = 3 \), \( P_4O_{10} = 1 \)