Nitromethane (CH₃NO₂), sometimes used as a fuel for drag racing, burns according to the following reaction:
4 CH₃NO₂ (l) + 7 O₂ (g) → 4 CO₂ (g) + 6 H₂O (g) + 4 NO₂ (g)   ∆Hº = – 2441.6 kJ
How much heat is released by burning 125.0 g of nitromethane (MW:61.044 g/mol)?

• A. 1.250 x 10³ kJ
• B. -1.250 x 10³ kJ
• C. 2.440 x 10³ kJ
• D. -2.440 x 10³ kJ

Answer: (B)

Verified step by step guidance

1. Determine the number of moles of nitromethane (CH3NO2) by using its mass and molar mass. Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Here, the mass is 125.0 g and the molar mass is 61.044 g/mol.
2. Calculate the moles of nitromethane: \( \text{moles of CH3NO2} = \frac{125.0 \text{ g}}{61.044 \text{ g/mol}} \).
3. Use the stoichiometry of the balanced chemical equation to find the relationship between moles of nitromethane and the enthalpy change (\( \Delta H^\circ \)). The reaction shows that 4 moles of CH3NO2 release 2441.6 kJ of heat.
4. Calculate the heat released for the moles of nitromethane you have. Use the proportion: \( \text{heat released} = \left( \frac{\text{moles of CH3NO2}}{4 \text{ moles}} \right) \times (-2441.6 \text{ kJ}) \).
5. Convert the calculated heat to the appropriate units and sign, ensuring it matches one of the given answer choices. Remember that the heat released should be a negative value, indicating an exothermic reaction.