How much volume of `CO_(2)` at S.T.P is liberated by the combustion of `100cm^(3)` of propane `(C_(3)H_(8))`?

Options

A. `100cm^(3)`
B. `200cm^(3)`
C. `300cm^(3)`
D. `400cm^(3)`

Explanation

To find out how much volume of \( CO_2 \) is liberated by the combustion of \( 100 \, cm^3 \) of propane \( (C_3H_8) \) at STP, we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of propane.
The combustion of propane can be represented by the following equation:

\[ C_3H_8 + O_2 \rightarrow CO_2 + H_2O \]

To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides.

1. **Carbon:** There are 3 carbons in \( C_3H_8 \), so we need 3 \( CO_2 \):
\[ C_3H_8 + O_2 \rightarrow 3CO_2 + H_2O \]

2. **Hydrogen:** There are 8 hydrogens in \( C_3H_8 \), so we need 4 \( H_2O \):
\[ C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O \]

3. **Oxygen:** Now we count the oxygen atoms. The products have \( 3 \times 2 = 6 \) from \( CO_2 \) and \( 4 \times 1 = 4 \) from \( H_2O \), totaling 10 oxygen atoms. Therefore, we need 5 \( O_2 \):
\[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \]

The balanced equation is:
\[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \]

### Step 2: Determine the number of moles of propane used.
To find the number of moles of propane in \( 100 \, cm^3 \):
- At STP, \( 1 \, mole \) of any gas occupies \( 22400 \, cm^3 \).
- Therefore, the number of moles of \( C_3H_8 \) in \( 100 \, cm^3 \) is calculated as follows:

\[ \text{Number of moles of } C_3H_8 = \frac{100 \, cm^3}{22400 \, cm^3/mol} = \frac{100}{22400} \, mol \]

### Step 3: Calculate the moles of \( CO_2 \) produced.
From the balanced equation, we see that 1 mole of \( C_3H_8 \) produces 3 moles of \( CO_2 \). Therefore, the moles of \( CO_2 \) produced from the moles of \( C_3H_8 \) is:

\[ \text{Moles of } CO_2 = 3 \times \text{Moles of } C_3H_8 = 3 \times \frac{100}{22400} \]

### Step 4: Convert moles of \( CO_2 \) to volume.
To find the volume of \( CO_2 \) produced, we use the fact that 1 mole of gas occupies \( 22400 \, cm^3 \):

\[ \text{Volume of } CO_2 = \text{Moles of } CO_2 \times 22400 \, cm^3/mol \]
\[ = 3 \times \frac{100}{22400} \times 22400 \]
\[ = 3 \times 100 = 300 \, cm^3 \]

### Final Answer:
The volume of \( CO_2 \) liberated by the combustion of \( 100 \, cm^3 \) of propane at STP is \( 300 \, cm^3 \).

---