To determine the amount of energy released when 59.7 grams of methane (CH₄) reacts with oxygen, we can follow these steps:
1. Identify the molar mass of methane (CH₄):
The molar mass of CH₄ is calculated from its constituent atoms:
[tex]\[ \text{Molar mass of CH₄} = (\text{molar mass of C}) + 4 \times (\text{molar mass of H}) \][/tex]
Given values:
- Molar mass of C (carbon) = 12.01 g/mol
- Molar mass of H (hydrogen) = 1.01 g/mol
So,
[tex]\[ \text{Molar mass of CH₄} = 12.01 + 4 \times 1.01 = 16.04\ \text{g/mol} \][/tex]
2. Calculate the number of moles of methane (CH₄):
[tex]\[ \text{Moles of CH₄} = \frac{\text{mass of CH₄}}{\text{molar mass of CH₄}} \][/tex]
Given:
- Mass of CH₄ = 59.7 g
- Molar mass of CH₄ = 16.04 g/mol
[tex]\[ \text{Moles of CH₄} = \frac{59.7\ \text{g}}{16.04\ \text{g/mol}} = 3.72\ \text{mol}\ (\text{to 3 significant figures}) \][/tex]
3. Determine the enthalpy change (ΔH) for the reaction:
The enthalpy change for the combustion of 1 mole of CH₄ is given as:
[tex]\[ ΔH = -890\ \text{kJ/mol} \][/tex]
4. Calculate the total energy released:
[tex]\[ \text{Energy released} = \text{moles of CH₄} \times \text{ΔH} \][/tex]
Given:
- Moles of CH₄ = 3.72 mol
- ΔH = -890 kJ/mol
[tex]\[ \text{Energy released} = 3.72\ \text{mol} \times -890\ \text{kJ/mol} \][/tex]
[tex]\[ \text{Energy released} = -3310\ \text{kJ} \ (\text{to 3 significant figures}) \][/tex]
Therefore, the combustion of 59.7 grams of methane releases -3310 kilojoules of energy.
