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

Options

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

Explanation

To determine the volume of \( CO_2 \) liberated by the combustion of \( 100 \, cm^3 \) of propane \( (C_3H_8) \) at Standard Temperature and Pressure (S.T.P.), we will 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 balanced equation:

\[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \]

### Step 2: Identify the volume ratio from the balanced equation.
From the balanced equation, we see that 1 volume of propane produces 3 volumes of \( CO_2 \). Therefore, the ratio of propane to \( CO_2 \) is:

\[ 1 \, C_3H_8 : 3 \, CO_2 \]

### Step 3: Calculate the volume of \( CO_2 \) produced from \( 100 \, cm^3 \) of propane.
Using the volume ratio, we can determine the volume of \( CO_2 \) produced from \( 100 \, cm^3 \) of propane:

\[ \text{Volume of } CO_2 = 100 \, cm^3 \times \frac{3 \, \text{volumes of } CO_2}{1 \, \text{volume of } C_3H_8} = 300 \, cm^3 \]

### Step 4: State the final answer.
Thus, the volume of \( CO_2 \) liberated by the combustion of \( 100 \, cm^3 \) of propane at S.T.P. is:

\[ \text{Volume of } CO_2 = 300 \, cm^3 \]

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