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 from its combustion, we can follow these steps:

### Step 1: Understand the given information
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 change in enthalpy (ΔH) for this reaction is given as -890.3 kJ. This means that when 1 mole of methane is combusted, 890.3 kJ of heat is released.

### Step 2: Calculate the amount of heat produced per gram of methane
First, we need to find out how much heat is produced per gram of methane. The molar mass of methane (CH₄) can be calculated as follows:
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.008 g/mol × 4 = 4.032 g/mol

Thus, the molar mass of CH₄ is:
\[ 12.01 + 4.032 = 16.042 \text{ g/mol} \]

Now, we can calculate the heat produced per gram of methane:
\[ \text{Heat per gram} = \frac{890.3 \text{ kJ}}{16.042 \text{ g}} \]

### Step 3: Calculate the heat produced per gram
Calculating this gives:
\[ \text{Heat per gram} = \frac{890.3}{16.042} \approx 55.5 \text{ kJ/g} \]

### Step 4: Determine how many grams are needed for 445.15 kJ
Now, we need to find out how many grams of methane are required to produce 445.15 kJ of heat:
\[ \text{Grams of CH}_4 = \frac{445.15 \text{ kJ}}{55.5 \text{ kJ/g}} \]

### Step 5: Perform the calculation
Calculating this gives:
\[ \text{Grams of CH}_4 \approx \frac{445.15}{55.5} \approx 8.02 \text{ g} \]

### Conclusion
Therefore, approximately 8.02 grams of methane are required to produce 445.15 kJ of heat.

### Final Answer
**8.02 grams of methane (CH₄) are required to produce 445.15 kJ of heat.**

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