To determine the amount of energy released when 59.7 grams of methane (CH[tex]\(_4\)[/tex]) reacts with oxygen, follow these detailed steps:
1. Determine the molar mass of methane (CH[tex]\(_4\)[/tex]):
Methane consists of one carbon atom and four hydrogen atoms.
- The atomic mass of carbon (C) is 12.01 g/mol.
- The atomic mass of hydrogen (H) is 1.008 g/mol.
Therefore, the molar mass of methane is:
[tex]\[
\text{Molar mass of CH}_4 = 12.01 + 4 \times 1.008 = 12.01 + 4.032 = 16.042 \text{ g/mol}
\][/tex]
2. Convert the mass of methane to moles:
Now, we need to convert the given mass of methane (59.7 grams) to moles using the molar mass:
[tex]\[
\text{Moles of CH}_4 = \frac{\text{Mass of CH}_4}{\text{Molar mass of CH}_4} = \frac{59.7 \text{ grams}}{16.042 \text{ g/mol}} \approx 3.721 \text{ moles}
\][/tex]
3. Use the given enthalpy change (ΔH):
The enthalpy change (ΔH) for the reaction is given as -890 kJ/mol. This means that for every mole of methane combusted, 890 kJ of energy is released.
4. Calculate the energy released:
To find the total energy released for 3.721 moles of methane, use the relationship:
[tex]\[
\text{Energy released} = \text{Moles of CH}_4 \times \Delta H = 3.721 \text{ moles} \times -890 \text{ kJ/mol} \approx -3312 \text{ kJ}
\][/tex]
5. Express the answer to three significant figures:
Hence, the amount of energy released when 59.7 grams of methane reacts with oxygen is:
[tex]\[
-3312 \text{ kJ}
\][/tex]
Therefore, the combustion of 59.7 grams of methane releases -3312 kJ of energy (to three significant figures).
