A gaseous hydrocarbon consumed 5 times its volume of oxygen as for combustion. The volume of `CO_(2)` produced in the reaction is thrice the volume of hydrocarbon under the same conditions
How many grams of water is produced by combustion of 0.1 mol of the givenn hydrocarbon?

Options

A. 7.2 gm
B. 3.6 gm
C. 14.4 gm
D. 1.8 gm

Answer

To solve the problem step by step, we will follow the combustion reaction of the hydrocarbon and derive the required information.

### Step 1: Identify the hydrocarbon and its combustion reaction
The combustion of a hydrocarbon can be represented by the general formula:
\[ \text{C}_x\text{H}_y + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]

### Step 2: Analyze the given information
1. The hydrocarbon consumes 5 times its volume of oxygen for combustion.
2. The volume of carbon dioxide produced is thrice the volume of the hydrocarbon.

Let’s denote the hydrocarbon as \( \text{C}_x\text{H}_y \).

### Step 3: Set up the equations based on the information
From the given information:
- The volume of oxygen consumed is \( 5 \) times the volume of the hydrocarbon.
- The volume of \( \text{CO}_2 \) produced is \( 3 \) times the volume of the hydrocarbon.

If we assume 1 mole of the hydrocarbon is combusted:
- Moles of \( \text{CO}_2 \) produced = \( 3 \)
- Moles of \( \text{O}_2 \) consumed = \( 5 \)

### Step 4: Relate the moles of \( \text{CO}_2 \) to the hydrocarbon
From the combustion reaction:
\[ \text{C}_x\text{H}_y + \frac{x + \frac{y}{4}}{5} \text{O}_2 \rightarrow x \text{CO}_2 + \frac{y}{2} \text{H}_2\text{O} \]

From the information:
- \( x = 3 \) (since \( \text{CO}_2 \) produced is three times the hydrocarbon)
- Therefore, \( y = 4 \) (derived from the equation \( \frac{3 + \frac{y}{4}}{5} = 5 \))

### Step 5: Determine the molecular formula of the hydrocarbon
Thus, the molecular formula of the hydrocarbon is \( \text{C}_3\text{H}_8 \) (propane).

### Step 6: Write the balanced combustion equation
The balanced equation for the combustion of propane is:
\[ \text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \]

### Step 7: Calculate the moles of water produced from 0.1 moles of hydrocarbon
From the balanced equation, 1 mole of \( \text{C}_3\text{H}_8 \) produces 4 moles of \( \text{H}_2\text{O} \).

Thus, for 0.1 moles of \( \text{C}_3\text{H}_8 \):
\[ \text{Moles of } \text{H}_2\text{O} = 0.1 \text{ moles of } \text{C}_3\text{H}_8 \times 4 = 0.4 \text{ moles of } \text{H}_2\text{O} \]

### Step 8: Convert moles of water to grams
The molar mass of water \( \text{H}_2\text{O} \) is approximately 18 g/mol.
\[ \text{Mass of } \text{H}_2\text{O} = \text{Moles} \times \text{Molar mass} = 0.4 \text{ moles} \times 18 \text{ g/mol} = 7.2 \text{ grams} \]

### Final Answer
Thus, the mass of water produced by the combustion of 0.1 moles of the hydrocarbon is **7.2 grams**.

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