A sample of gaseous hydrocarbon occupying `1.12 "litre"` at `NTP`, when completely burnt in air produced `2.2 g CO_(2)` and `1.8 g H_(2)O`. Calculate the weight of hydrocarbon taken and the volume of `O_(2)` at `NTP` required for its combustion.

Options

A. `CH_4`
B. `C_2 H_6`
C. `C_3 H_8`
D. `C_4 H_10`

Answer

To solve the problem, we need to determine the weight of the hydrocarbon taken and the volume of O₂ required for its combustion. Let's break it down step by step.

### Step 1: Determine the number of moles of CO₂ and H₂O produced.
- The molar mass of CO₂ = 12 (C) + 16*2 (O) = 44 g/mol.
- The molar mass of H₂O = 2*1 (H) + 16 (O) = 18 g/mol.

Using the given masses:
- Moles of CO₂ produced = mass / molar mass = 2.2 g / 44 g/mol = 0.05 moles.
- Moles of H₂O produced = mass / molar mass = 1.8 g / 18 g/mol = 0.1 moles.

### Step 2: Relate the moles of hydrocarbon (CxHy) to the products.
From the combustion reaction:
\[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \]

- Each mole of hydrocarbon produces x moles of CO₂ and y/2 moles of H₂O.
- From the moles of CO₂ produced (0.05 moles), we can say:
\[ x = 0.05 \text{ moles of CO₂} \]
- From the moles of H₂O produced (0.1 moles), we can say:
\[ \frac{y}{2} = 0.1 \text{ moles of H₂O} \Rightarrow y = 0.2 \text{ moles} \]

### Step 3: Calculate the values of x and y.
From the above equations:
- We have x = 1 (since 0.05 moles of CO₂ corresponds to 1 mole of hydrocarbon).
- We have y = 4 (since 0.1 moles of H₂O corresponds to 4 hydrogen atoms).

### Step 4: Identify the hydrocarbon.
The hydrocarbon can be identified as \( C_1H_4 \) which is methane (CH₄).

### Step 5: Calculate the weight of the hydrocarbon taken.
- Molar mass of CH₄ = 12 (C) + 4 (H) = 16 g/mol.
- Since we have 0.05 moles of CH₄:
\[ \text{Weight of hydrocarbon} = \text{moles} \times \text{molar mass} = 0.05 \text{ moles} \times 16 \text{ g/mol} = 0.8 \text{ g} \]

### Step 6: Calculate the volume of O₂ required for combustion.
From the balanced equation:
\[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \]
- 1 mole of CH₄ requires 2 moles of O₂ for complete combustion.
- Therefore, 0.05 moles of CH₄ will require:
\[ 0.05 \text{ moles of CH₄} \times 2 = 0.1 \text{ moles of O₂} \]

At NTP, 1 mole of gas occupies 22.4 liters:
- Volume of O₂ required = 0.1 moles × 22.4 L/mol = 2.24 L.

### Final Answers:
- Weight of hydrocarbon taken = **0.8 g**
- Volume of O₂ required = **2.24 L**

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