The mass of `P_(4)O_(10)` produced if 440 gm `P_(4)S_(3)` is mixed with 384 gm
of `O_(2)` is

Options

A. 568gm
B. 426 gm
C. 284 gm
D. 396 gm

Answer

To find the mass of \( P_4O_{10} \) produced when 440 g of \( P_4S_3 \) is mixed with 384 g of \( O_2 \), we will follow these steps:

### Step 1: Write the Balanced Chemical Equation
The balanced chemical reaction for the combustion of \( P_4S_3 \) in oxygen is:

\[ P_4S_3 + O_2 \rightarrow P_4O_{10} + SO_2 \]

Balancing the equation gives us:

\[ 4P_4S_3 + 8O_2 \rightarrow 4P_4O_{10} + 3SO_2 \]

### Step 2: Calculate Molar Masses
Next, we calculate the molar masses of the reactants and products.

- For \( P_4S_3 \):
\[ \text{Molar mass of } P_4S_3 = (31 \times 4) + (32 \times 3) = 124 + 96 = 220 \, \text{g/mol} \]

- For \( O_2 \):
\[ \text{Molar mass of } O_2 = 16 \times 2 = 32 \, \text{g/mol} \]

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

### Step 3: Calculate Moles of Reactants
Now, we calculate the number of moles of each reactant.

- Moles of \( P_4S_3 \):
\[ \text{Moles of } P_4S_3 = \frac{440 \, \text{g}}{220 \, \text{g/mol}} = 2 \, \text{moles} \]

- Moles of \( O_2 \):
\[ \text{Moles of } O_2 = \frac{384 \, \text{g}}{32 \, \text{g/mol}} = 12 \, \text{moles} \]

### Step 4: Determine the Limiting Reagent
Using the stoichiometry from the balanced equation, we find the limiting reagent.

From the balanced equation:
- 1 mole of \( P_4S_3 \) reacts with 2 moles of \( O_2 \).

Thus, for 2 moles of \( P_4S_3 \):
\[ \text{Required moles of } O_2 = 2 \times 2 = 4 \, \text{moles} \]

Since we have 12 moles of \( O_2 \), \( P_4S_3 \) is the limiting reagent.

### Step 5: Calculate Moles of \( P_4O_{10} \) Produced
From the balanced equation:
- 1 mole of \( P_4S_3 \) produces 1 mole of \( P_4O_{10} \).

Thus, 2 moles of \( P_4S_3 \) will produce:
\[ \text{Moles of } P_4O_{10} = 2 \, \text{moles} \]

### Step 6: Calculate the Mass of \( P_4O_{10} \)
Finally, we calculate the mass of \( P_4O_{10} \) produced:
\[ \text{Mass of } P_4O_{10} = \text{Moles} \times \text{Molar mass} = 2 \, \text{moles} \times 284 \, \text{g/mol} = 568 \, \text{g} \]

### Final Answer
The mass of \( P_4O_{10} \) produced is **568 grams**.

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