On combustion, one mole of a hydrocarbon gives 2 moles of carbon dioxide and two moles of water and a large quantity of heat is produced. The hydrocarbon is

Options

A. methane
B. ethane
C. ethene
D. ethyne

Answer

To determine the hydrocarbon that produces 2 moles of carbon dioxide and 2 moles of water upon combustion, we can follow these steps:

### Step 1: Write the General Combustion Reaction
The general combustion reaction for a hydrocarbon (C_xH_y) can be represented as:
\[ \text{C}_x\text{H}_y + O_2 \rightarrow CO_2 + H_2O + \text{Heat} \]

### Step 2: Set Up the Combustion Equation
According to the question, one mole of the hydrocarbon produces 2 moles of carbon dioxide (CO₂) and 2 moles of water (H₂O). Thus, we can write the equation as:
\[ \text{C}_x\text{H}_y + O_2 \rightarrow 2 \text{CO}_2 + 2 \text{H}_2O \]

### Step 3: Balance the Equation
From the products, we can see that:
- 2 moles of CO₂ means there are 2 carbon atoms in the hydrocarbon (C_x).
- 2 moles of H₂O means there are 4 hydrogen atoms in the hydrocarbon (H_y).

So, we can conclude:
- \( x = 2 \) (2 carbon atoms)
- \( y = 4 \) (4 hydrogen atoms)

This gives us the formula for the hydrocarbon:
\[ \text{C}_2\text{H}_4 \] (which is ethene).

### Step 4: Determine the Oxygen Requirement
Now, we need to balance the oxygen atoms. The products have:
- 2 moles of CO₂ contribute \( 2 \times 2 = 4 \) oxygen atoms.
- 2 moles of H₂O contribute \( 2 \times 1 = 2 \) oxygen atoms.

Total oxygen needed = \( 4 + 2 = 6 \) oxygen atoms.

Since each O₂ molecule contains 2 oxygen atoms, we need:
\[ \frac{6}{2} = 3 \text{ moles of } O_2 \]

### Step 5: Write the Final Balanced Equation
Now we can write the complete balanced equation for the combustion of ethene:
\[ \text{C}_2\text{H}_4 + 3 O_2 \rightarrow 2 CO_2 + 2 H_2O + \text{Heat} \]

### Conclusion
The hydrocarbon that produces 2 moles of carbon dioxide and 2 moles of water upon combustion is **ethene (C₂H₄)**.

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