The enthalpy of combustion of methane at `25^(@)C` is 890 kJ. The heat liberated when 3.2 g of methane is burnt in air is

Options

A. 445 kJ
B. 278 kJ
C. `-890 kJ`
D. 178 kJ

Explanation

To find the heat liberated when 3.2 g of methane (CH₄) is burnt in air, we can follow these steps:

### Step 1: Determine the molar mass of methane (CH₄).
The molar mass of methane is calculated as follows:
- Carbon (C) = 12 g/mol
- Hydrogen (H) = 1 g/mol × 4 = 4 g/mol
- Molar mass of CH₄ = 12 g/mol + 4 g/mol = 16 g/mol

### Step 2: Calculate the number of moles of methane in 3.2 g.
To find the number of moles of methane, use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{3.2 \text{ g}}{16 \text{ g/mol}} = 0.2 \text{ moles} \]

### Step 3: Calculate the heat liberated using the enthalpy of combustion.
The enthalpy of combustion of methane is given as 890 kJ per mole. Therefore, the heat liberated when 0.2 moles of methane is burnt can be calculated as:
\[ \text{Heat liberated} = \text{Number of moles} \times \text{Enthalpy of combustion} \]
Substituting the values:
\[ \text{Heat liberated} = 0.2 \text{ moles} \times 890 \text{ kJ/mol} = 178 \text{ kJ} \]

### Final Answer:
The heat liberated when 3.2 g of methane is burnt in air is **178 kJ**.

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