Given the thermochemical equation CH_4(g) + 2 O_2(g) → CO_2(g) + 2 H_2O(l); ΔH = -890 kJ, what is the enthalpy change (ΔH) when 530 g of methane (CH_4) is combusted?

• A. -16,600 kJ
• B. -890 kJ
• C. -29,400 kJ
• D. -29.4 kJ

Answer: (C)

Verified step by step guidance

1. Identify the given thermochemical equation and its enthalpy change: \(\mathrm{CH_4(g) + 2\ O_2(g) \rightarrow CO_2(g) + 2\ H_2O(l)}\) with \(\Delta H = -890\ \mathrm{kJ}\). This means that burning 1 mole of methane releases 890 kJ of energy.
2. Calculate the number of moles of methane combusted using its given mass. Use the molar mass of methane (\(\mathrm{CH_4}\)), which is calculated as: \$12.01\ \mathrm{g/mol}\( (C) + \)4 \times 1.008\ \mathrm{g/mol}\( (H) = approximately \)16.04\ \mathrm{g/mol}\(. Then, find moles by \)n = \frac{\text{mass}}{\text{molar mass}}$.
3. Use the stoichiometry of the reaction to relate moles of methane to enthalpy change. Since the enthalpy change is given per mole of methane combusted, multiply the number of moles calculated by \(\Delta H\) per mole: \(\Delta H_{total} = n \times (-890\ \mathrm{kJ/mol})\).
4. Make sure to keep track of the sign of \(\Delta H\) (negative in this case) because it indicates an exothermic reaction (heat is released).
5. Express the final enthalpy change for the combustion of 530 g of methane in kJ, which will be the product of the number of moles and the enthalpy change per mole.