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. First, determine the number of moles of nitromethane (CH3NO2) in 125.0 g. Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). The molar mass of nitromethane 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. Next, use the stoichiometry of the reaction to determine the heat released. The reaction shows that 4 moles of CH3NO2 release 2441.6 kJ of heat.
4. Set up a proportion to find the heat released by the moles of nitromethane calculated: \( \text{heat released} = \left( \frac{\text{moles of CH3NO2}}{4 \text{ moles}} \right) \times 2441.6 \text{ kJ} \).
5. Finally, ensure the sign of the heat released is negative, as the reaction is exothermic, meaning heat is released.