`10mL` of gaseous hydrocarbon on combustion gives `40ml` of `CO_(2)(g)` and `50mL` of `H_(2)O` (vapour) The hydrocarbon is .

Options

A. `C_(4)H_(5)`
B. `C_(8)H_(10)`
C. `C_(4)H_(8)`
D. `C_(4)H_(10)`

Explanation

To determine the formula of the hydrocarbon from the given combustion data, we can follow these steps:

### Step 1: Understand the combustion reaction
The combustion of a hydrocarbon (CₓHᵧ) in the presence of oxygen (O₂) produces carbon dioxide (CO₂) and water (H₂O). The general reaction can be expressed as:
\[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \]

### Step 2: Analyze the volumes of products formed
From the problem, we know:
- Volume of CO₂ produced = 40 mL
- Volume of H₂O produced = 50 mL

### Step 3: Relate the volumes of products to the number of moles
According to Avogadro's law, equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Therefore, we can relate the volumes of CO₂ and H₂O to the moles of carbon (C) and hydrogen (H) in the hydrocarbon.

1. **For CO₂**:
Each mole of CO₂ corresponds to one mole of carbon in the hydrocarbon. Thus:
\[ \text{Moles of C} = \frac{40 \text{ mL}}{22.4 \text{ L/mol}} = \frac{40}{22400} \text{ mol} \]
Since we are using volumes directly, we can say:
\[ \text{Moles of C} = \frac{40}{10} = 4 \text{ moles of C} \]

2. **For H₂O**:
Each mole of H₂O corresponds to two moles of hydrogen in the hydrocarbon. Thus:
\[ \text{Moles of H} = 2 \times \frac{50 \text{ mL}}{22.4 \text{ L/mol}} = 2 \times \frac{50}{22400} \text{ mol} \]
Again, using volumes directly:
\[ \text{Moles of H} = 2 \times \frac{50}{10} = 10 \text{ moles of H} \]

### Step 4: Write the empirical formula
From the calculations:
- Number of carbon atoms (C) = 4
- Number of hydrogen atoms (H) = 10

Thus, the empirical formula of the hydrocarbon is:
\[ C_4H_{10} \]

### Final Answer
The hydrocarbon is **butane (C₄H₁₀)**.

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