To determine how much energy is released when 59.7 grams of methane (CH[tex]\(_4\)[/tex]) reacts with oxygen, follow these steps:
1. Identify the molar mass of methane (CH[tex]\(_4\)[/tex]):
The molar mass of methane is:
[tex]\( \text{C: } 12.01 \text{ g/mol} \)[/tex]
[tex]\( \text{H: } 1.01 \times 4 = 4.04 \text{ g/mol} \)[/tex]
Therefore, the total molar mass of CH[tex]\(_4\)[/tex] is [tex]\( 12.01 + 4.04 = 16.05 \text{ g/mol} \)[/tex].
2. Convert the given mass of methane to moles:
Given that the mass of methane is 59.7 grams, we can find the number of moles by dividing the mass by the molar mass:
[tex]\[
\text{Moles of CH\(_4\)} = \frac{59.7 \text{ grams}}{16.05 \text{ g/mol}} = 3.722 \text{ moles}
\][/tex]
3. Calculate the energy released:
According to the enthalpy change ([tex]\( \Delta H \)[/tex]) provided in the combustion reaction, there is an energy release of -890 kJ per mole of methane.
Thus, the total energy released is:
[tex]\[
\text{Energy released} = \text{Moles of CH\(_4\)} \times \Delta H = 3.722 \text{ moles} \times (-890 \text{ kJ/mol})
\][/tex]
This calculation yields:
[tex]\[
\text{Energy released} = -3312.531 \text{ kJ}
\][/tex]
Rounding to three significant figures, the energy released is -3310 kJ.
Therefore, the combustion of 59.7 grams of methane releases -3310 kilojoules of energy.
