What mass of `P_(4) O_(10)` will be produced by the combustion of `2.0 g` of `P_(4)` with `2.0 g` of `O_(2)`

Options

A. `1.04 g`
B. `0.52 g`
C. `2.04 g`
D. `3.04 g`

Explanation

To solve the problem of determining the mass of \( P_4O_{10} \) produced by the combustion of \( 2.0 \, g \) of \( P_4 \) with \( 2.0 \, g \) of \( O_2 \), we can follow these steps:

### Step 1: Write the balanced chemical equation

The combustion of phosphorus (\( P_4 \)) in oxygen (\( O_2 \)) can be represented by the following balanced equation:

\[ P_4 + 5 O_2 \rightarrow P_4O_{10} \]

### Step 2: Calculate the moles of \( P_4 \) and \( O_2 \)

To find the number of moles, we use the formula:

\[ \text{Moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]

**For \( P_4 \):**
- Molar mass of \( P_4 = 4 \times 31 \, g/mol = 124 \, g/mol \)
- Given mass of \( P_4 = 2.0 \, g \)

\[ \text{Moles of } P_4 = \frac{2.0 \, g}{124 \, g/mol} = \frac{2.0}{124} \approx 0.0161 \, moles \]

**For \( O_2 \):**
- Molar mass of \( O_2 = 2 \times 16 \, g/mol = 32 \, g/mol \)
- Given mass of \( O_2 = 2.0 \, g \)

\[ \text{Moles of } O_2 = \frac{2.0 \, g}{32 \, g/mol} = \frac{2.0}{32} = 0.0625 \, moles \]

### Step 3: Determine the limiting reagent

From the balanced equation, we see that 1 mole of \( P_4 \) reacts with 5 moles of \( O_2 \).

To find out how much \( O_2 \) is needed for the available \( P_4 \):

\[ \text{Required moles of } O_2 = 5 \times \text{moles of } P_4 = 5 \times 0.0161 \approx 0.0805 \, moles \]

We have \( 0.0625 \, moles \) of \( O_2 \), which is less than \( 0.0805 \, moles \). Therefore, \( O_2 \) is the limiting reagent.

### Step 4: Calculate the moles of \( P_4O_{10} \) produced

From the balanced equation, 5 moles of \( O_2 \) produce 1 mole of \( P_4O_{10} \).

Using the moles of \( O_2 \):

\[ \text{Moles of } P_4O_{10} = \frac{0.0625 \, moles \, O_2}{5} = 0.0125 \, moles \]

### Step 5: Calculate the mass of \( P_4O_{10} \)

To find the mass of \( P_4O_{10} \), we need its molar mass:

- Molar mass of \( P_4O_{10} = 4 \times 31 + 10 \times 16 = 124 + 160 = 284 \, g/mol \)

Now, we can calculate the mass:

\[ \text{Mass of } P_4O_{10} = \text{moles} \times \text{molar mass} = 0.0125 \, moles \times 284 \, g/mol = 3.55 \, g \]

### Final Answer

The mass of \( P_4O_{10} \) produced is approximately **3.55 grams**.

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