The volume of `CO_2` prodcued by the combination of 40 ml of gaseous acetone in excess of oxygen is :

Options

A. 40 ml
B. 80 ml
C. 60 ml
D. 120 ml

Answer

To find the volume of \( CO_2 \) produced by the combustion of 40 ml of gaseous acetone in excess oxygen, we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of acetone.
The combustion of acetone (\( C_3H_6O \)) in the presence of oxygen (\( O_2 \)) produces carbon dioxide (\( CO_2 \)) and water (\( H_2O \)). The balanced equation is:

\[ C_3H_6O + 4 O_2 \rightarrow 3 CO_2 + 3 H_2O \]

### Step 2: Analyze the stoichiometry of the reaction.
From the balanced equation, we can see that 1 mole of acetone produces 3 moles of \( CO_2 \). This means that for every 1 volume of acetone, 3 volumes of \( CO_2 \) are produced.

### Step 3: Use the volume of acetone to find the volume of \( CO_2 \).
Given that we have 40 ml of acetone, we can calculate the volume of \( CO_2 \) produced using the stoichiometric ratio:

\[ \text{Volume of } CO_2 = 3 \times \text{Volume of acetone} \]

Substituting the given volume of acetone:

\[ \text{Volume of } CO_2 = 3 \times 40 \, \text{ml} = 120 \, \text{ml} \]

### Step 4: Conclusion
Thus, the volume of \( CO_2 \) produced by the combustion of 40 ml of gaseous acetone in excess of oxygen is **120 ml**.