The heat evolved in the combustion of methane is given by the following equation :
`CH_(4)(g)+2O_(2)(g)to CO_(2)(g)+2H_(2)O(l) , Delta H = -890.3 kJ`
How many grams of methane would be required to produce 445.15 kJ of heat of combustion ?

Options

A. 4 g
B. 8 g
C. 12 g
D. 16 g

Answer

To solve the problem of how many grams of methane (CH₄) are required to produce 445.15 kJ of heat during combustion, we can follow these steps:

### Step 1: Understand the Reaction
The combustion of methane is represented by the equation:
\[ \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \]
The heat evolved (ΔH) for this reaction is -890.3 kJ, indicating that 890.3 kJ of heat is released when 1 mole of methane is burned.

### Step 2: Determine the Molar Mass of Methane
The molar mass of methane (CH₄) can be calculated as follows:
- Carbon (C) has a molar mass of 12 g/mol.
- Hydrogen (H) has a molar mass of 1 g/mol, and there are 4 hydrogen atoms in methane.

Thus, the molar mass of methane is:
\[ 12 \, \text{g/mol} + (4 \times 1 \, \text{g/mol}) = 16 \, \text{g/mol} \]

### Step 3: Calculate the Amount of Heat Released per Gram of Methane
Since 1 mole of methane (16 g) releases 890.3 kJ of heat, we can calculate the heat released per gram of methane:
\[ \text{Heat per gram} = \frac{890.3 \, \text{kJ}}{16 \, \text{g}} \]
\[ \text{Heat per gram} = 55.65 \, \text{kJ/g} \]

### Step 4: Calculate the Grams of Methane Required for 445.15 kJ
To find out how many grams of methane are needed to produce 445.15 kJ of heat, we can set up the following proportion:
\[ \text{Grams of methane} = \frac{445.15 \, \text{kJ}}{55.65 \, \text{kJ/g}} \]

Calculating this gives:
\[ \text{Grams of methane} = \frac{445.15}{55.65} \approx 8 \, \text{g} \]

### Conclusion
Therefore, approximately 8 grams of methane are required to produce 445.15 kJ of heat during combustion.

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