To determine how much energy is released when 59.7 grams of methane reacts with oxygen, let's break down the steps needed to solve the problem:
1. Identify Given Values:
- The enthalpy change ([tex]\(\Delta H\)[/tex]) for the reaction: [tex]\(-890 \text{ kJ/mol}\)[/tex]
- The mass of methane reacting: [tex]\(59.7 \text{ grams}\)[/tex]
- The molar mass of methane ([tex]\(\text{CH}_4\)[/tex]): [tex]\(16.04 \text{ g/mol}\)[/tex]
2. Calculate the Number of Moles of Methane:
- Number of moles is found by dividing the mass of the substance by its molar mass.
[tex]\[
\text{Moles of methane} = \frac{\text{mass of methane}}{\text{molar mass of methane}} = \frac{59.7 \text{ grams}}{16.04 \text{ g/mol}} \approx 3.721 \text{ moles}
\][/tex]
3. Calculate the Energy Released:
- Energy released is found by multiplying the number of moles by the enthalpy change ([tex]\(\Delta H\)[/tex]).
[tex]\[
\text{Energy released} = \text{moles of methane} \times \Delta H = 3.721 \text{ moles} \times (-890 \text{ kJ/mol}) \approx -3312.531 \text{ kJ}
\][/tex]
4. Express the Answer to Three Significant Figures:
- The energy released should be expressed correctly to three significant figures.
[tex]\[
\text{Energy released} \approx -3312.531 \text{ kJ} \rightarrow -3312.531 \approx -3310 \text{ kJ}
\][/tex]
Thus, the combustion of 59.7 grams of methane releases [tex]\(\boxed{-3310}\)[/tex] kilojoules of energy.
